Sainavle

Expansion Of Gases Long Answer Type Questions

Question 1. In case of volume expansion of a gas, mention of both the pressure and the temperature is necessary, whereas for expansion of solids and liquids, only the temperature is mentioned. Why?
Answer:

The effect of change in pressure on the volume of a solid or of a liquid is practically insignificant because of compactness of molecules. But, inter-molecular attraction in gases is weak so change in pressure significantly changes the volume of a gas.

All substances, solid, liquid or gas, expand on increase in temperature in general. Hence, the mention of pressure along with temperature is necessary in case of volume expansion of a gas.

Question 2. When a balloon is inflated both its volume and its pressure increase. Is there any violation of Boyle’s law?
Answer:

According to Boyles law, the volume of a gas varies inversely with pressure at constant temperature, only if the mass of the gas is fixed. During the inflation of a balloon, the mass of a gas is increasing. So there is no violation of Boyle’s law.

Question 3. Unlike a liquid, there is no coefficient of apparent expansion in case of a gas—why? Or, During the expansion of a liquid, volume expansion of the container is taken into account, but not for a gas—why?
Answer:

When a liquid in a container is heated, both the liquid and the container expand. The coefficient of expansion of any solid is less than that of any liquid but not negligible.

• So the expansion of the solid is not neglected. Thus in case of liquids we get two coefficients of expansion. One is the coefficient of apparent expansion and the other is the coefficient of real expansion.
• In apparent expansion the expansion of the container is ignored, whereas in real expansion, the coefficient of expansion of the container is added with the coefficient of apparent expansion.
• A mass of a gas is also heated by heating the container. This is similar to the heating of a liquid. But for the same rise in temperature, the expansion of a gas is nearly 100 times more than that of the container.
• Unless a very accurate measurement is required, expansion of the container is neglected. Hence, no apparent expansion needs to be considered in case of a gas.

Question 4. Two identical spherical bulbs contain air and are connected by a short horizontal glass tube. The tube contains a short mercury thread in it. Temperatures of the two bulbs are 0°C and 20°C respectively. If the temperature of each bulb is increased by 10°C, what will be the change in the position of the mercury thread?
Answer:

Given

Two identical spherical bulbs contain air and are connected by a short horizontal glass tube. The tube contains a short mercury thread in it. Temperatures of the two bulbs are 0°C and 20°C respectively. If the temperature of each bulb is increased by 10°C

The mercury thread shift⇒ s towards the bulb at higher temperature. As the mercury thread is in equilibrium before increasing the temperature, the initial pressure on both sides of the thread should be the same, say p.

On heating, the equilibrium is disturbed and the pressures change to p₁ and p₂ respectively for the bulbs at (0+ 10)°C or 10°C, and (20 + 10)°C or 30°C respectively.

Applying Charles’ law for the first bulb, \(\frac{p_1}{p}=\frac{T_1}{T}=\frac{273+10}{273}=\frac{283}{273}=1.037\)

and for the second bulb \(\frac{p_2}{p}=\frac{T_2}{T}=\frac{273+30}{273+20}=\frac{303}{293}=1.034 \)

⇒ \(\frac{p_1}{p_2}=\frac{283 \times 293}{273 \times 303}=1.0024\)

As p₁ is greater than p₂, the thread shifts towards the bulb at higher temperature.

Question 5. To define the coefficient of expansion of gases, the initial volume or pressure is always taken at 0°C. But for the coefficients of expansion of solids and liquids, the initial temperature need not be taken as 0°C. Why?
Answer:

The values of the coefficients of expansion of solids and liquids are very small. Hence, in this case, the volume at any temperature can be taken as the initial volume. The coefficient of volume expansion of a gas is relatively higher.

So if we consider volumes or pressures at different temperatures as the initial volume or pressure, the coefficient of expansion differs considerably. Hence, to define the coefficient of expansion of gases, the initial volume or pressure should always be taken at 0°C.

Question 6. Air pressure in a car tyre increases during driving. Explain why.
Answer:

Air pressure in a car tyre increases during driving.

Due to friction between the road and the car tyre, the temperature of air inside the tyre increases reasonably. This increases the air pressure during driving.

Thermal Expansion vs. Gas Expansion Questions

Question 7. The expansion of a gas follows the condition pV² = constant. Show that such an expansion causes cool- ingof the gas.
Answer:

Given

The expansion of a gas follows the condition pV² = constant.

Assume that the initial volume of the gas is V₁ at pressure p₁ and at temperature T₁. After the expansion, the corresponding quantities are V₂, p₂ and T₂ (say). From ideal gas equation,

⇒ \(\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \text { or, } \frac{p_1 V_1}{p_2 V_2}=\frac{T_1}{T_2}\)

or, \(\frac{p_1 V_1^2}{p_2 V_2^2}=\frac{T_1 V_1}{T_2 V_2}\) multiplying both sides by \(\frac{V_1}{V_2}\)

As the condition is, \(p_1 V_1^2=p_2 V_2^2\), we have

⇒ \(\frac{V_1 T_1}{V_2 T_2}=1 \text { or, } \frac{T_1}{T_2}=\frac{\dot{V}_2}{V_1}\)

As the gas expands, \(V_2>V_1\) therefore \(\frac{T_1}{T_2}>1\) or, \(T_2<T_1\).

∴ The gas cools down upon expansion.

Question 8. For a fixed mass of a gas at constant volume, draw p-t°C and p-TK graphs. How can the value of, absolute zero be obtained from the 1st graph?
Answer:

The graphs are shown in Figures respectively for 3 different volumes.

The graphs are straight lines in nature.

In the p-t°C graph, the three lines corresponding to the three volumes, converge at a point on the negative t axis, where pressure is zero. This point gives the value of absolute zero as per definition.

In the p-TK graph, the three straight lines pass through the origin, showing p = 0 when T = 0.

Question 9. For a fixed mass of a gas at constant pressure, draw V-t° C and V- T K graphs. How can the value of absolute zero be obtained from the 1st graph?
Answer:

The graphs are shown in Figures respectively for three different pressures.

The graphs are straight lines in nature.

In the V-t°C graph shown, the three lines corresponding to the three pressures, converge at a point on the negative t-axis, where volume is zero. This point gives the value of absolute zero as per definition.

In the V-T K graph the three straight lines pass through the origin, showing V = 0 when T = 0.

Question 10. Determine the value of universal gas constant JR and gas constant k for lg of air. (At STP the density of air =1.293 g · L^-1 and that of mercury = 13.6 g · cm^-3)
Answer:

Pressure p – 76 x 13.6 x 980 dyn · cm^-2 and temperature = 0°C = 273 K

Volume of 1 mol of any gas at STP = 22.4 L = 22400 cm³

∴ R = \(\frac{p V}{T}=\frac{76 \times 13.6 \times 980 \times 22400}{273}\)

= \(8.31 \times 10^7 \mathrm{erg} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

Now, volume of 1 g air,

v = \(\frac{1}{1.293} l=\frac{1000}{1.293} \mathrm{~cm}^3\)

∴ k = \(\frac{p v}{T}=\frac{76 \times 13.6 \times 980 \times 1000}{1.293 \times 273}\)

= \(0.287 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Question 11. A given mass of an Ideal gas is heated in a vessel. The same amount of gas is then heated by keeping it in a larger vessel. Assume that the volumes of both vessel remain the same during heating. What will be the nature of the pressure-temperature (p- T) graphs in the two cases?
Answer:

Given

A given mass of an Ideal gas is heated in a vessel. The same amount of gas is then heated by keeping it in a larger vessel. Assume that the volumes of both vessel remain the same during heating.

In each case the graph is a straight line, For a Fixed mass of gas, when the volume is less, the pressure increases with temperature at a higher rate.

So the slope of the p- T graph for the smaller container is greater than that for the bigger container.

Alternative method: For a fixed mass of an ideal gas, \(p V=k T \text { or, } p=\frac{k}{V} T \text {. }\)

pV = kT or, p = k/V T

If V is fixed, this relation resembles, y = mx.

So the p – T graph is a straight line passing through the origin.

Also, the slope k/V is higher for a smaller V. So the graph for the smaller vessel has a higher slope.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Sample Problems on Ideal Gas Expansion

Question 12. Draw p- T graphs for masses m and 2m of the same gas, when heated in a container of constant volume. Interpret the slopes.
Answer:

The desired graphs are shown in Figure.

The equation of state for mass m of a gas of molecular weight M is, \(p V=\frac{m}{M} R T \quad \text { or } p=\frac{m R}{M V} T\)

Given, V = constant; so for any fixed mass, p = constant x T.

Thus, the p- T graph is a straight line passing through the origin.

Also, the slope \(\frac{m R}{M V}\) is higher for a higher m.

∴ The slope of the graph for mass 2 m is more than that for mass m.

For any temperature the value of p in the first case will be half of second case. \(\left(B C=\frac{1}{2} A C\right)\)

Question 13. Shows the V- T graph for a fixed mass of an ideal gas at pressures p₁ and p₂. Can you infer from the graph whether p₁ is greater than p₂?
Answer:

We have, pV= nRT,

or, V = \(\frac{n R}{p} T .\)

This shows that the V- T graph is a straight line passing through the origin, for any fixed pressure p (n = constant for a fixed mass).

The slope \(\frac{n R}{p}\) is greater for a smaller value of p.

In Figure, the graph for p₂ has a higher slope.

So, p₂ < p₁.

Thus, p₁ is greater than p₂.

Question 14. In a faulty barometer, some air is occupying the space over mercury column. How can the air pressure be correctly determined with this faulty barometer?
Answer:

Let the length of mercury column = h₁, and that of air column over mercury =l₁.

Now the barometer tube is raised a little still keeping the open end dipped into mercury. Let in this case, the length of mercury column = h₂, and that of air column over mercury = l₁.

If the true reading of the air pressure is H, and area of the cross-section of the tube is α, then from Boyle’s law

⇒ \(\left(H-h_1\right) l_1 \alpha=\left(H-h_2\right) l_2 \alpha \quad \text { or, } H=\frac{h_1 l_1-h_2 l_2}{l_1-l_2}\)

Hence, measuring h₁, h₂, l₁, l₂ the correct atmospheric pressure can be found out.

Question 15. A container filled with oxygen is taken to the moon’s surface from the earth. How will the volume and pressure of the gas change when the container is

1. A rubber balloon,
2. A steel cylinder?

Answer:

1. Atmospheric pressure on moon’s surface is practically zero. As pressure of oxygen inside the rubber balloon is high, the gas will expand in volume and finally the balloon will burst.

2. Volume of the gas in the steel cylinder would remain unchanged. There will be no effect of zero atmospheric pressure.

But the pressure of the gas would change due to a different temperature of the surroundings.

Question 16. What is meant by specific gas constant? Is the value of this constant same for all gases?
Answer:

Value of \(\frac{p V}{T}\) for 1 g of a gas is the specific heat constant of that gas. Value of the constant is different for different gases because of the difference in molecular weight M, since \(\frac{p V}{T}=n R=\frac{1}{M} R \text {. }\)

Question 17. Equal number of hydrogen and helium molecules are kept in two identical gas jars at the same temperature. What will be the ratio of the pressures of the gases in the two jars?
Answer:

Given

Equal number of hydrogen and helium molecules are kept in two identical gas jars at the same temperature.

Pressure is the same in both the jars. As per Avaga- dro’s law, all gases of equal volume contain the same number of molecules under identical values of temperature and pressure. In this example, volume, temperature and number of molecules are the same. So, the pressure will be equal in the two jars and the ratio is 1:1.

Question 18. A gas container contains 1 mol of O₂ gas (specific molar mass 32) at pressure p and temperature T. In a similar container one mol of He gas (specific molar mass 4) is kept at temperature 2 T. What is the pressure of this He gas?
Answer:

Given

A gas container contains 1 mol of O₂ gas (specific molar mass 32) at pressure p and temperature T. In a similar container one mol of He gas (specific molar mass 4) is kept at temperature 2 T.

Since number of moles and volume are the same for both the gases, using pV = nRT, it can be said that the pressure is directly proportional to the absolute temperature. Since the temperature of the second container is double, pressure will also be double, i.e., 2p.

Question 19. Same ideal gas is kept in two containers A and B fitted with frictionless pistons. Volume and temperature of the gas in both containers are the same. m_A and m_B are the masses of the gas in A and B respectively. The volume of the gases in the two containers are changed to 2 V keeping their temperature constant.

Corresponding changes in pressure in A and B are Δp and 1.5 Δp. Find the ratio of the masses of the gas kept in A and in B.

Answer:

Given

Same ideal gas is kept in two containers A and B fitted with frictionless pistons. Volume and temperature of the gas in both containers are the same. m_A and m_B are the masses of the gas in A and B respectively. The volume of the gases in the two containers are changed to 2 V keeping their temperature constant.

Corresponding changes in pressure in A and B are Δp and 1.5 Δp.

Volumes of the gases in both the containers become double. So, according to Boyle’s law, the corresponding pressures of the gases become half of their initial pressures.

∴ Initial pressure in A and B are,

p_A = 2Δp and p_B = 2 x 1.5Δp = 3Δp

∴ \(\frac{p_A}{p_B}=\frac{2}{3} \text { or, } \frac{n_A \frac{R T}{V}}{n_B \frac{R T}{V}}=\frac{2}{3}\)

or, \(\frac{m_A \frac{R T}{M V}}{m_B \frac{R T}{M V}}=\frac{2}{3} \text { or, } \frac{m_A}{m_B}=\frac{2}{3}\)

Question 20. An ideal gas is found to obey a gas law, Vp² = constant Initial temperature and volume of the gas are T and V respectively. If the gas expands to a volume 2 V, what will be the effect on temperature?
Answer:

Given

An ideal gas is found to obey a gas law, Vp² = constant Initial temperature and volume of the gas are T and V respectively. If the gas expands to a volume 2 V,

Here Vp² = constant…..(1)

For an ideal gas pV = RT or \(p=\frac{R T}{V}\)

Substituting for p, equation (1) gives

⇒ \(\frac{V \times R^2 T^2}{V^2}=\text { constant or, } \frac{T^2}{V}=\text { constant }\)

When the volume of the gas expands to 2V, suppose the temperature is T’.

∴ \(\frac{T^2}{V}=\frac{T^{\prime 2}}{2 V} \text { or, } \frac{T^{\prime 2}}{T^2}=2 \text { or, } T^{\prime}=T \sqrt{2} \text {. }\)

Question 21. Shows the p- T graphs for a fixed mass of an ideal gas at volumes V₁ and V₂. Can it be concluded from the graphs that V₁ is greater than V₂?
Answer:

∴ \(p V=n R T=\frac{m}{M} R T\)

or, p = \(\frac{m R}{M V} T\) [M= molecular weights]

So p- T graphs are straight lines passing through the origin, for any fixed V. The slope \(\frac{m R}{M V}\) is smaller for a greater value of V.

The slope for V₁ is smaller than that for V₂. So, V₁>V₂.

Question 22. An ideal gas is Initially at temperature T and volume V. Its volume increases by dV due to an increase in temperature dT, while pressure remains constant. Here \(\gamma=\frac{1}{V} \frac{d V}{d T}\).What will be the nature of the graph between γ and T1
Answer:

In case of an ideal gas,

pV= RT or, \(V=\frac{R T}{p}\)

When pressure remains constant, we have \(\frac{d V}{d T}=\frac{R}{p}\)

∴ \(\gamma=\frac{1}{V} \frac{d V}{d T}=\frac{1}{V} \frac{R}{p}=\frac{R}{R T}=\frac{1}{T}\)

So, the graph γ-T will be a rectangular hyperbola.

Question 23. An ideal gas is initially at pressure p and volume V. Its pressure is increased by dp, so that its volume decreases by dV, while the temperature remains constant. Here \(\beta=-\frac{1}{V} \frac{d V}{d p}\). What will be the nature of the graph between β and p?
Answer:

Given

An ideal gas is initially at pressure p and volume V. Its pressure is increased by dp, so that its volume decreases by dV, while the temperature remains constant.

In case of an ideal gas, pV = RT

If the temperature remains constant, p V = constant

∴ pdV+ Vdp = 0

or, \(\frac{d V}{d p}=-\frac{V}{p}\)

∴ \(\beta=-\frac{1}{V} \frac{d V}{d p}=\frac{1}{p}\)

i. e., βp = 1 = constant So, the graph β-p will be a rectangular hyperbola.

Question 24. Pressure coefficient of a gas is \(\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\) Explain.
Answer:

Pressure coefficient of a gas is \(\frac{1}{273}{ }^{\circ} \mathrm{C}^{-1}\)

It means that when the temperature increases 1°C, at constant volume the increase in pressure of the gas of fixed mass per unit pressure at 0°C is equal to 1/273 of the original pressure of that gas at °C.

Expansion Of Gases Multiple Choice Questions And Answers

WBBSE Class 11 Thermal Expansion MCQs

Question 1. Both the volume and the pressure of a definite mass of gas are observed to increase. This is possible when the temperature of the gas

1. Remains the same
2. Decreases
3. Increases
4. First decreases, then increases

Answer: 3. Increases

Question 2. The value of the specific gas constant of hydrogen is

1. \(4.16 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)
2. \(0.26 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)
3. \(4.80 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)
4. \(5.16 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Answer: 1. \(4.16 \times 10^7 \mathrm{erg} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

Question 3. At constant pressure, if the temperature of a gas is increased then its density

1. Remains the same
2. Decreases
3. Increases
4. Increases or decreases depending on the nature of the gas

Answer: 2. Decreases

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. The isothermal (temperature = constant) graph of a gas is its

1. P-V graph
2. p-1/V graph
3. pV-p graph
4. pV-V graph

Answer: 1. P-V graph

Conceptual Questions on Thermal Expansion for Class 11

Question 5. At constant pressure the volume of a definite mass of gas changes with its temperature

1. Non-linearly
2. Linearly
3. In the form of a rectangular hyperbola
4. None of the above

Answer: 2. Linearly

Question 6. pV-p graph of an ideal gas is

1. Parallel to p-axis
2. Parallel to pv-axis
3. Not parallel to any axis
4. Rectangular hyperbolic

Answer: 1. Parallel lo p-axis

Question 7. A vessel contains 1 mol of O₂ gas (specific molar mass 32) at a temperature T. Pressure of this gas is p. In another identical vessel, 1 mol of He gas (specific molar mass 4) is kept at a temperature 2T. The pressure of this gas will be

1. p/8
2. p
3. 2p
4. 8p

Answer: 3. 2p

Question 8. The unit of pV in the equation pV = RT is

1. N · m^-1
2. J
3. J · K^-1
4. None of these

Answer: 2. J

Practice MCQs on Linear and Volume Expansion

Question 9. Two gases having the same pressure p, volume V and temperature T are mixed with each other. If the volume and temperature of the mixture are V and T respectively, then the value of pressure will be

1. 2p
2. p
3. p/2
4. 4p

Answer: 1. 2p

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 10. Compared to that of solids and liquids, value of the coefficient of volume expansion of gases

1. Is same
2. Is comparatively greater
3. Is comparatively less

Answer: 2. Is comparatively greater

Question 11. If the coefficients of volume expansion of a solid, a liquid and a gas are γ_s, γ_l and γ_g respectively then

1. For Different Solids, Liquids And Gases The Values Of γ_s, γ_l And γ_g Are Different
2. For Different Solids And Liquids The Values Of γ_s And γ_l Are Different, But For All Gases The Value Of γ_g Is The Same
3. For Different Solids The Values Of γ_s Are Different But For All Liquids The Value Of γ_l And For All Gases The Value Of γ_g Are The Same
4. For All Solids, Liquids AndGases The Values Of γ_s, γ_l And γ_g Respectively AreThe Same

Answer: 2. For Different Solids And Liquids The Values Of γ_s And γ_l Are Different, But For All Gases The Value Of γ_g Is The Same

Question 12. Coefficients of volume expansion of solids, liquids and gases are respectively γ_s, γ_l And γ_g Usually

1. \(\gamma_s<\gamma_l<\gamma_g\)
2. \(\gamma_s>\gamma_l>\gamma_g\)
3. \(\gamma_l<\gamma_s<\gamma_g\)
4. \(\gamma_l>\gamma_g>\gamma_s\)

Answer: 1. \(\gamma_s<\gamma_l<\gamma_g\)

Key MCQs on Coefficient of Thermal Expansion

Question 13. The volume of a gas at STP is 150 cm³, At constant volume, the pressure of the gas becomes 850 mmllg at a temperature of 25°C. The pressure coefficient of that gas is

1. 4.73 x 10^-3 °C^-1
2. 5.73 x 10^{-3 °C-1}
3. 6.73 x10^{-3 °C-1}
4. 1 °C^-1

Answer: 1. 4.73 x 10^-3 °C^-1

Question 14. While determining the volume coefficient of a gas, the initial volume is taken as its volume at

1. 273°C
2. 0°C
3. 100°C
4. 27°C

Answer: 2. 0°C

Question 15. Volume coefficient and pressure coefficient are equal in case of

1. Ideal gas
2. Real gas
3. Hydrogen
4. Inert gases

Answer: 1. Ideal gas

Question 16. An ideal gas is expanding such that pT² = constant. The coefficient of volume expansion of the gas is

1. \(\frac{1}{T}\)
2. \(\frac{2}{T}\)
3. \(\frac{3}{T}\)
4. \(\frac{4}{T}\)

Answer: 3. \(\frac{3}{T}\)

Question 17. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and volume. The mass of the gas in A is mA and the same in B is m_B. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2 V. The change in the pressure in A and B are found to be Δp and 1.5 Δp respectively. Then

1. 4m_A = 9m_B
2. 2m_A = 3m_B
3. 3m_A = 2m_B
4. 9m_A = 4m_B

Answer: 3. 3m_A = 2m_B

Question 18. Concerning the three quantities—pressure p, density d, and absolute temperature T—the gas equation can be written as

1. \(\frac{p_1}{T_1 d_1}=\frac{p_2}{T_2 d_2}\)
2. \(\frac{p_1 T_1}{d_1}=\frac{p_2 T_2}{d_2}\)
3. \(\frac{p_1 d_1}{T_2}=\frac{p_2 d_2}{T_1}\)
4. \(\frac{p_1 d_1}{T_1}=\frac{p_2 d_2}{T_2}\)

Answer: 1. \(\frac{p_1}{T_1 d_1}=\frac{p_2}{T_2 d_2}\)

Question 19. At a pressure p, volume V, and temperature T, the equation of state for 5 g of oxygen will be [R = molar gas constant]

1. \(p V=\frac{5}{32} R T\)
2. \(p V=5 R T\)
3. \(p V=\frac{5}{2} R T\)
4. \(p V=\frac{5}{16} R T\)

Answer: 1. \(p V=\frac{5}{32} R T\)

Sample Questions on Applications of Thermal Expansion

Question 20. When an air bubble rises from the bottom of a lake to the surface, its radius is doubled. Atmospheric pressure is equal to the pressure of a water column of height H. Depth of the lake is

1. H
2. 2H
3. 7H
4. 8H

Answer: 3. 7H

In this type of question, more than one option are correct.

Question 21. In the thermal expansion of an ideal gas

1. There is no change in the temperature of the gas
2. There is no change in the internal energy of the gas
3. The work done by the gas is equal to the heat supplied to the gas
4. The work done by the gas is equal to the change in its internal energy

Answer:

1. There is no change in the temperature of the gas
2. There is no change in the internal energy of the gas
3. The work done by the gas is equal to the heat supplied to the gas

WBBSE Class 11 Practice Tests on Thermal Properties

Question 22. From the following statements concerning ideal gas at any given temperature T, select the correct one(s).

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases
2. The coefficient of pressure expansion at constant volume is the same for all ideal gases
3. The coefficient of pressure expansion and volume expansion are not equal for any ideal gas
4. The coefficient of pressure expansion and volume expansion are equal for all ideal gases

Answer:

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases

2. The coefficient of pressure expansion at constant volume is the same for all ideal gases

4. The coefficient of pressure expansion and volume expansion are equal for all ideal gases

Question 23. Which of the following statements is true?

1. The density of a gas is proportional to the absolute temperature at a constant pressure
2. The density of a gas is inversely proportioned to the absolute temperature at constant pressure
3. The density of a gas is proportional to the pressure at a constant temperature
4. The density of a gas is inversely proportional to the pressure at a constant temperature

Answer:

2. The density of a gas is inversely proportioned to the absolute temperature at constant pressure

3. The density of a gas is proportional to the pressure at a constant temperature.

Class 11 Physics – Calorimetry Multiple Choice Questions And Answers

Calorimetry MCQs for Class 11

Question 1. The amount of heat required by 1 g of a substance for its 1 °C rise in temperature is called its

1. Specific heat
2. Thermal capacity
3. Water equivalent
4. Latent heat

Answer: 1. Specific heat

Question 2. SI unit of heat is

1. cal
2. kcal
3. J
4. W

Answer: 3. J

Question 3. Mean calorie means

1. The amount of heat required to increase the temperature of 1 g water from 0°c to 1°c
2. The amount of heat required to increase the temperature of 1 g water from 50°c to 51 °c
3. The amount of heat required to increase the temperature of 1 g water from 14.5°c to 15.5°c
4. 1/100 part of the amount of heat required to increase the temperature of 1 g water from 0°C to 100°C

Answer: 4. 1/100 part of the amount of heat required to increase the temperature of 1 g water from 0°C to 100°C

Question 4. If the amount of heat required by mass m of a substance for a rise t in temperature be H, then

1. \(t \propto m H\)
2. \(t \propto \frac{H}{m}\)
3. \(t \propto \frac{m}{H}\)
4. \(t \propto \frac{1}{m H}\)

Answer: 2. \(t \propto \frac{H}{m}\)

Question 5. cal · g^-1 · °C^-1 is the unit of

1. Specific heat capacity
2. Thermal capacity
3. Water equivalent
4. Latent heat

Answer: 1. Specific heat capacity

Question 6. Which of the following substances has the highest specific heat?

1. Mercury
2. Water
3. Iron
4. Diamond

Answer: 2. Water

MCQ Practice on Thermal Properties of Matter

Question 7. Due to the higher specific heat of water compared to other solids and liquids

1. Water warms up quickly but cools down slowly
2. Water cools down quickly but warms up slowly
3. Water warms up or cools down slowly
4. Water warms up or cools down quickly

Answer: 3. Water warms up or cools down slowly

Question 8. To cool down the engine of a car, water is used in the radiator because

1. Water is easily available
2. Water does not cause any harm to the radiator
3. The viscosity of water is much less
4. The specific heat of water is high

Answer: 4. Specific heat of water is high

Question 9. During boiling of water at 100°C, what will be its specific heat?

1. Zero
2. 0.5
3. 1
4. Infinite

Answer: 4. Infinite

Question 10. If the temperature of 1 g of water is raised by 1°C for what initial temperature the heat gained is equal in magnitude to the mean calorie?

1. 0°C
2. 14.5°C
3. 15°C
4. 15.5°C

Answer: 2. 14.5°C

Question 11. An alternative name of ‘mean calorie’ is

1. 0°- 100°C cal
2. 14.5°C cal
3. 15°C cal
4. 15.5°C cal

Answer: 3. 15°C cal

Calorimetry Problems with Solutions

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 12. The temperature range and pressure, at which the heat required by 1 g water for 1° C rise in its temperature is called 1 cal, is

1. 3.5°C to 4.5°C, 76 cm of Hg
2. 13.5°C to 14.5°C, 76 mm of Hg
3. 14.5°C to 15.5°C, 760 mm of Hg
4. 98.5°C to 99.5°C, 760 mm of Hg

Answer: 3. 14.5°C to 15.5°C, 760 mm of Hg

Question 13. 10 g of water is supplied with 420 J of energy. The rise in temperature of water is

1. 1°C
2. 4.2°C
3. 10°C
4. 32°C

Answer: 3. 10°C

Question 14. If the specific heat of a substance is infinite, it means

1. Heat is given out
2. Heat is taken in
3. No change in temperature takes place whether heat is taken in or given out
4. All of the above

Answer: 3. No change in temperature takes place whether heat is taken in or given out

Question 15. Calorimeters are made of which of the following?

1. Glass
2. Metal
3. Wood
4. Either 1 or 3

Answer: 2. Metal

Interactive Calorimetry MCQ Tests

Question 16. Mass m of a substance requires an amount H ofheatfor a rise r in temperature. If the specific heat of the substance be s, then its water equivalent is

1. \(H m s_w\)
2. \(m s_w t\)
3. \(\frac{m s}{s_w}\)
4. \(\frac{s t}{s_w}\)

Answer: 3. \(\frac{m s}{s_w}\)

Question 17. Thermal capacity of a body of mass 10 g is 8 cal · °C^-1. The specific heat of the material of the body is

1. 0.8
2. 1.25
3. 0.4
4. 0.1

Answer: 1. 0.8

Question 18. If the specific heat of copper is 0.1 cal · g^-1 · °C^-1, then water equivalent of a copper calorimeter of mass 0.4 kg is

1. 40 g
2. 4000 g
3. 200 g
4. 4g

Answer: 1. 40 g

Question 19. The ratio of the radii of two spheres made of the same material is 1:4. The ratio of their thermal capacities is

1. 1/4
2. 1/32
3. 1/2
4. 1/4

Answer: 1. 1/4

Heat Transfer and Calorimetry MCQs

Question 20. The ratio of the densities of two materials is 5 : 6 and of their specific heats is 3:5. The ratio of their thermal capacities per unit volume will be

1. 1:2
2. 2:1
3. 3:2
4. 2:3

Answer: 1. 1:2

Question 21. When two bodies at different temperatures are brought in contact, the basic principle of calorimetry (heat lost = heat gained) can not be applied if

1. The two bodies do not mix well with each other
2. Any of the bodies undergo a change of state
3. Specific beats of the two bodies are widely different
4. Chemical reaction occurs between the bodies

Answer: 4. Chemical reaction occurs between the bodies

Question 22. If no change of state occurs, which of the following quantities is not required for the calculation of heat lost or heat gained?

1. Mass
2. Density
3. Specific Heat
4. Change In Temperature

Answer: 2. Density

Question 23. The ratio of specific heats of two liquids is 1: 2. If the two liquids at different temperatures are mixed in the ratio 2 : 3 of their masses, then what will be the ratio of h changes in their temperatures?

1. 1:3
2. 1:6
3. 3:1
4. 6:1

Answer: 3. 3:1

Question 24. The specific heat of aluminium is more than that of copper. TWo spheres of equal masses made of these two metals are immersed in a hot liquid. In equilibrium

1. The temperature of aluminium sphere will be greater.
2. The temperature of both spheres will be equal
3. The temperature of copper sphere will be greater
4. None of the above

Answer: 2. The temperature of both the spheres will be equal

Question 25. A liquid of mass M and specific heat S is at a temperature 2t. If another liquid of thermal capacity 1.5 times, at a temperature of | is added to it, the resultant temperature will be

1. 4/3 t
2. t
3. t/2
4. 2/3 t

Answer: 2. t

In tills type of questions more than one options are correct.

Question 26. Thermal capacity of a body depends on

1. The heat given
2. The temperature raised
3. The mass of the body
4. The material of the body

Answer:

3. The mass of the body

4. The material of the body

Question 27. The specific heat of a substance can be

1. Finite
2. Infinite
3. Zero
4. Negative

Answer:

1. Finite
2. Infinite
3. Zero

Question 28. When two samples at different temperatures are mixed, the temperature of the mixture can be

1. Lesser than lower or greater than higher temperature
2. Equal to lower or higher temperature
3. Greater than lower but lesser than higher temperature
4. Average of lower and higher temperature

Answer:

2. Equal to lower or higher temperature

3. Greater than lower but lesser than higher temperature

4. Average of lower and higher temperature

Calorimetry Long Answer Type Questions And Answers

Calorimetry Questions and Answers for Class 11 WBCHSE

Question 1. Equal masses of milk and water are taken in two identical kettles and are heated by the same source. The rate of rise of the temperature of milk is found to be higher than that of water. Explain.
Answer:

Given

Equal masses of milk and water are taken in two identical kettles and are heated by the same source. The rate of rise of the temperature of milk is found to be higher than that of water.

Since both the kettles are heated by the same source, in the same interval of time amount of heat (H) absorbed is the same. Now if the rate of absorption of heat \(\left(\frac{H}{\text { time }}\right)\) and mass (m) are fixed, then from the equation H = mst we get,

st ∝ constant

∴ \(t \propto \frac{1}{s}\)

Now, since specific heat of milk is less than that of water, rate of increase in temperature of the milk is greater than that of water of the same mass.

Question 2. Milk and water of equal mass are taken in two similar vessels at room temperature. Both of them are to be heated from room temperature to a certain higher temperature. Which will take less heat and why?
Answer:

Given

Milk and water of equal mass are taken in two similar vessels at room temperature. Both of them are to be heated from room temperature to a certain higher temperature.

Let mass of milk = m = mass of water

Increase in temperature of milk = t = increase in temperature of water

Let specific heats of milk and water be s_m and s_w.

Heat absorbed by milk, H_m = ms_mt

Heat absorbed by water, H_w = ms_wt

∴ \(s_m<s_w, \quad H_m<H_w\)

So, less heat is required to warm milk.

Question 3. 1 kg of iron at 100°C melts more Ice than 1 kg of lead at 100°C. Explain why.
Answer:

Given

1 kg of iron at 100°C melts more Ice than 1 kg of lead at 100°C.

Specific heat of iron is more than that of lead. Hence, for the same fall in temperature, iron supplies more heat to the ice than lead of the same mass does. Hence, iron melts relatively more ice.

Question 4. What is the advantage of taking water in hot water bottles?
Answer:

The advantage of taking water in hot water bottles are

Water has a specific heat higher than everything but ammonia. Hence, a fixed mass of water gains more heat than any other liquid of the same mass for the same rise in temperature.

Consequently, it loses more heat during cooling. This heat is used for fomentation. So a ot water bottle remains effective for a long period of time.

Key Questions on Calorimetry for Class 11

Question 5. Two copper spheres of the some external radius, one solid but the other hollow, are heated up to a certain temperature and then are allowed to cool under similar conditions. Which sphere will cool faster?
Answer:

Given

Two copper spheres of the some external radius, one solid but the other hollow, are heated up to a certain temperature and then are allowed to cool under similar conditions.

The hollow sphere will cool faster.

As both are made of copper, the specific heat s is the same.

Now, H = mst, or t = H/ms, where t = rate of fall of temperature. As the mass of the hollow sphere is less, t will be higher for it. So it will cool faster.

Question 6. State whether the fundamental law of calorimetry is applicable in the following cases:

1. Sugar is added to water taken in a calorimeter,
2. A chemical reaction occurs between a solid and a liquid in a calorimeter,
3. Calorimeter is kept open in air.

Answer:

While dissolving in water, sugar will absorb the necessary heat of the solution. If it is not taken into consideration, the fundamental law of calorimetry cannot be applied.

• During a chemical reaction, heat is evolved or absorbed. If this heat is not taken into consideration, the fundamental law of calorimetry cannot be applied.
• If the calorimeter is kept open in air, heat exchange occurs with the surroundings. Hence, the fundamental law of calorimetry will not be applicable.

Question 7. Two small spherical balls of the same mass, made of copper and of lead are heated up to the same tern- perature and are placed on a thick wax slab. What will happen and what can we conclude from it?
Answer:

Given

Two small spherical balls of the same mass, made of copper and of lead are heated up to the same tern- perature and are placed on a thick wax slab.

The copper ball sinks in the wax slab more than the lead ball does, because the rate of melting of wax due to the copper ball is greater. From this we conclude that the heat loss of the copper ball is greater than that of the lead ball. But, the mass and the decrease in temperatures of both the balls are equal. Therefore, from the equation H = mst we conclude, the specific heat (s) of copper is higher.

Question 8. If two bodies of equal mass but of different materials are supplied equal amounts of heat, which one will have higher rise in temperature?
Answer:

Given, Heat H  = mst i.e,  temperature rise, t = H/ms

If two bodies of equal mass but of different materials are supplied equal amounts of heat,

Since for the two bodies, heat supplied H and mass m are equal, \(t \propto \frac{1}{s}\)

Hence, the body with comparatively less specific heat will have a higher rise in temperature.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 9. 100 g of water and 100 g of iron are heated up to the same temperature and are separately added to 50 g of water at a lower temperature, kept in two identical vessels. Compare the temperature changes in the two vessels.
Answer:

Given

100 g of water and 100 g of iron are heated up to the same temperature and are separately added to 50 g of water at a lower temperature, kept in two identical vessels.

Temperatures of both the vessels will increase. The specific heat of iron is much less than that of water (s_water = 1 and s_iron = 0.117). The initial temperatures of the given specimens of 100 g of water and 100 g of iron are equal.

But, due to its higher specific heat, water will lose more heat than iron will. Hence, the final temperature of the vessel in which 100 g of hot water is added will be higher.

Practice Questions on Calorimetry for Class 11

Question 10. When a hot body heats up a cold body, is the temperature change of both the bodies the same? Explain the answer.
Answer:

The change in temperature of the two bodies would differ. Heat lost by the hot body or gained by the cold body is equal to the product of

1. Mass
2. Specific heat and
3. Change in temperature.

Hence, the body having higher mass and higher specific heat will have less change in temperature for the same amount of heat transfer.

Question 11. Why calorimeters are made of metal (mostly copper) instead of glass?
Answer:

Due to higher conductivity of metals, in a metallic calorimeter thermal equilibrium is achieved very fast. Also specific heat of metals is lower than that of glass.

• That is why the water equivalent of a metallic calorimeter is lower than that of a glass calorimeter of the same mass. Therefore, for the same rise in temperature heat gained by metallic calorimeter is less than that gained by glass calorimeter.
• As a result the final temperature of the mixture will be higher and will result less error in measurement.
• Copper has the highest conductivity (other than silver) among all the metals, and copper is much cheaper than silver, that is why copper is mostly used as the material of a calorimeter.

Question 12. The temperature of a furnace is more than 500°C. Discuss a method of measuring the temperature of the furnace with the help of a thermometer graduated up to 100°C.
Answer:

Given

The temperature of a furnace is more than 500°C

Let the temperature of the furnace = t°C. A small metal piece of mass m and specific heat s is put into the furnace. Obviously, the melting point of the metal should be greater than the temperature of the furnace so that it does not melt.

Now, water of mass M is taken in a vessel of water equivalent W and its temperature is recorded. Let the temperature be  t₁(< 100°C).

Now the hot metal piece is dropped in that water. As a result, a small quantity of water is vapourised which is negligible. After a while, when the metal piece and water reach thermal equilibrium, the final temperature is measured by the thermometer.

If that temperature is t₂°C, heat lost by the metal piece = ms(t- t₂) and heat gained by the vessel and water = (W+ M)s_w(t₂ – t₁), where s_w is the specific heat of water.

From calorimetric principle, ms(t-t₂) = (W+ M)s_w(t₂ – t₁)

Value of l can be calculated from this equation if values of all the other quantities are known. Thus, measuring temperatures t₁ and t₂ with the help of a thermometer graduated up to 100°C, a much greater temperature t can he determined.

Examples of Calorimetry Questions with Answers

Question 13. The diameter of an iron sphere and the length of the side of an iron cube are equal. Initially they are at the same temperature. If they take the same amount of heat, whose final temperature will be higher?
Answer:

Given

The diameter of an iron sphere and the length of the side of an iron cube are equal. Initially they are at the same temperature. If they take the same amount of heat,

If x be the diameter of the sphere, its volume is, \(V_1=\frac{4}{3} \pi\left(\frac{x}{2}\right)^3\)

The length of the side of the cube is x.

The volume of the cube is, V₂ = x³

∴ \(\frac{V_1}{V_2}=\frac{\frac{4}{3} \pi \frac{x^3}{8}}{x^3}=\frac{\pi}{6}<1 \quad therefore V_1<V_2\)

Since the volume of the cube is greater than that of the sphere, the mass of the cube is also greater. So, if an equal amount of heat is taken by the cube, its rise in temperature will be smaller, i.e., the rise in temperature of the sphere will be higher.

WBCHSE Class 11 Physics On Elasticity Short Questions And Answers

Question 1. A spring is cut into two equal pieces. What is the spring the constant of each part if the spring constant of the original spring is k,
Solution:

Let us consider that the spring elongates by x when a force F is applied on it. So, the force constant of the spring, k = F/x.

Now, if the spring is cut into two equal parts, then on the application of the same force F, each part of the spring will elongate by x/2.

The force constant each part, \(k^{\prime}=\frac{F}{\frac{x}{2}}=\frac{2 F}{x}=2 k\)

Question 2. A spring having spring constant k is cut into two parts in the ratio 1:2. Find the spring constants of the two parts.
Solution:

Let the initial length of the spring be x.

The spring constant of a particular spring is inversely proportional to its length.

∴ kx = constant.

When the spring is cut into two parts in the ratio 1:2, the length of the two parts are x/3 and 2x/3 respectively.

⇒\(k_1 \frac{x}{3}=k x \text { or, } k_1=3 k\)

and \(k_2 \cdot \frac{2 x}{3}=k x \text { or, } k_2=\frac{3 k}{2}\)

Question 3. The length of a metal wire is L₁. when the tension is T₁ and L₂ when the tension is T₂ The unstretched length of the wire is

1. \(\frac{L_1+L_2}{2}\)
2. \(\sqrt{L_1 L_2}\)
3. \(\frac{T_2 L_1-T_1 L_2}{T_2-T_1}\)
4. \(\frac{T_2 L_1+T_1 L_2}{T_2+T_1}\)

Answer:

Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}\)

or, strain = \(\frac{\text { stress }}{\text { strain }}\)

If the length of the wire is L₀ when there is no tension in the string, then in the first case, stress = \(\frac{T_1}{\alpha}\) and strain = \(\frac{L_1-L_0}{L_0}\)

[a = area of cross-section = constant (approximately)]

So, \(\frac{L_1-L_0}{L_0}=\frac{T_1}{\alpha Y} \quad \text { or, } \frac{1}{\alpha Y}=\frac{1}{T_1}\left(\frac{L_1}{L_0}-1\right)\)

Similarly in the second case, \(\frac{1}{\alpha Y}=\frac{1}{T_2}\left(\frac{L_2}{L_0}-1\right)\)

So, \(\frac{1}{T_1}\left(\frac{L_1}{L_0}-1\right)=\frac{1}{T_2}\left(\frac{L_2}{L_0}-1\right)\)

or, \(\frac{1}{L_0}\left(\frac{L_1}{T_1}-\frac{L_2}{T_2}\right)=\frac{1}{T_1}-\frac{1}{T_2}\)

or, \(\frac{1}{L_0} \frac{T_2 L_1-T_1 L_2}{T_1 T_2}=\frac{T_2-T_1}{T_1 T_2}\)

∴ \(L_0=\frac{T_2 L_1-T_1 L_2}{T_2-T_1}\)

The option 3  is correct

Question 4. A liquid of bulk modulus k is compressed by applying an external pressure such that its density increases by 0.04%. The pressure applied to the liquid is

1. k/10000
2. k/10000
3. 1000k
4. 0.01k

Answer:

k = \(\frac{p}{\frac{\Delta V}{V}}\)

or, \(p=k \times \frac{\Delta V}{V}=k \times \frac{\Delta \rho}{\rho}=k \times 0.01 \%=\frac{k}{10000}\)

The option 1 is correct.

Question 5. The stress along the length of a rod (with a rectangular cross section) is 1% of the Young’s modulus of its material. What is the approximate percentage of change in its volume? (Poisson’s ratio of the material of the rod is 0.3)

1. 3%
2. 1%
3. 0.7%
4. 0.4%

Answer:

Let, the volume of the rod, V = xyz, and Young’s modulus of its material of the rod = Y

Now, \(\frac{F}{A}=Y \times 1 \%\)

or, \(Y \times \frac{\Delta x}{x}=\frac{Y}{100}\)

or, \(\frac{\Delta x}{x}=0.01\)

∴ \(\frac{\Delta V}{V}=\frac{\Delta x}{x}+\frac{\Delta y}{y}+\frac{\Delta z}{z}\)

= \(\frac{\Delta x}{x}-\sigma \frac{\Delta x}{x}-\sigma \frac{\Delta x}{x}\) ……..(1)

[Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}=\frac{\frac{\Delta z}{z}}{\frac{\Delta x}{x}}\)]

The negative symbol in equation (1) implies that, as length increases due to stress, the value of y and z decreases simultaneously.

∴ From equation (1),

∴ \(\frac{\Delta V}{V}=0.01-2 \times 0.3 \times 0.01=0.004=0.4 \%\)

The option 4 is correct.

WBBSE Class 11 Elasticity Short Questions and Answers

Question 6. When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants. The work done in stretching the unstretched rubber band by L isothermal

1. \(a L^2+b L^3\)
2. \(\frac{1}{2}\left(a L^2+b L^3\right)\)
3. \(\frac{a L^2}{2}+\frac{b L^3}{3}\)
4. \(\frac{1}{2}\left(\frac{a L^2}{2}+\frac{b L^3}{3}\right)\)

Answer:

⇒ \(\int d W=\int F d l\)

W = \(\int_0^L a x d x+\int_0^L b x^2 d x=\frac{a L^2}{2}+\frac{b L^3}{3}\)

The option 3 is correct.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 7. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of

1. 9
2. 1/9
3. 81
4. 1/81

Answer:

According to the question, \(\frac{V_f}{V_i}=(9)^3\)

So, \(\frac{m_f}{m_i}=(9)^3\)

Also, \(\frac{A_f}{A_i}=(9)^2\)

Stress = \(\frac{\text { force }}{\text { area }}=\frac{m \times g}{A}\)

∴ \(\frac{S_f}{S_i}=\frac{m_f}{m_i} \times \frac{A_i}{A_f}=(9)^3 \times \frac{1}{(9)^2}=9\)

The option 1 is correct.

Question 8. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating it. The temperature should be raised by

1. \(\frac{P}{3 \alpha K}\)
2. \(\frac{P}{\alpha K}\)
3. \(\frac{3 \alpha}{P K}\)
4. \(3 P K \alpha\)

Answer:

Bulk modulus, K= \(\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume]\)

If we bring back the cube to its original size by increasing the temperature Δt,

⇒ \(\Delta V=V \cdot \gamma \Delta t\)

or, \(\Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 k \alpha}\)

The option (1) is correct.

Question 9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area floats on the surface of the liquid, covering the entire cross-section of the cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (dr/r) is

1. \(\frac{m g}{3 K a}\)
2. \(\frac{m g}{K a}\)
3. \(\frac{K a}{m g}\)
4. \(\frac{K a}{3 m g}\)

Answer:

Bulk modulus,  K = \(-V \frac{d p}{d V}\)

or, \(-\frac{d V}{V}=\frac{d p}{K}\)

or, \(\frac{-3 d r}{r}=\frac{\frac{m g}{a}}{K}\left[because V=\frac{4}{3} \pi r^3\right]\)

or, \(\frac{d r}{r}=-\frac{m g}{3 K a} \quad therefore\left|\frac{d r}{r}\right|=\frac{m g}{3 K a}\)

The option 1 is correct

Question 10. The copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?

1. Δl versus 1/l
2. Δl versus l²
3. Δl versus 1/l²
4. Δl versus l

Answer:

Y = \(\frac{F l}{A \Delta l} \text { or, } \Delta l=\frac{F l}{A Y}=\frac{F l^2}{V Y}\)

∴ \(\Delta l \propto l^2\)

The option 2 is correct.

Short Answer Questions on Stress and Strain

Question 11. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pal^-1 and the density of water is 10³kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

1. 0.8 x 10^-2
2. 1.0 x 10^-2
3. 1.2 x 10^-2
4. 1.4×10^-2

Answer:

Due to AP amount of increase in pressure, there is AV
amount of compression in volume V.

So, fractional compression = \(\frac{\Delta V}{V}\)

and compressibility, K = \(\frac{1}{V} \frac{\Delta V}{\Delta P}\)

Now consider the magnitude, \(\frac{\Delta V}{V}=K \Delta P\)

Here, ΔP = hρg = 2700 x 10³ x 10 Pa [taking g = 10m/s²]

Hence, fractional compression, \(\frac{\Delta V}{V} =\left(45.4 \times 10^{-11}\right) \times\left(2700 \times 10^3 \times 10\right)\)

= \(1.226 \times 10^{-2}\)

The option 3 is correct.

Question 12. The density of a metal at normal pressure is p. Its density when it is subjected to an excess pressure p is p’. If B is the bulk modulus of the metal, the ratio of \(\frac{e^{\prime}}{\rho}\)

1. \(1+\frac{B}{p}\)
2. \(\frac{1}{1-\frac{p}{B}}\)
3. \(1+\frac{p}{B}\)
4. \(\frac{1}{1+\frac{P}{B}}\)

Answer:

Volume strain = change in pressure = p

Initial volume, V = \(\frac{M}{\rho}\)

Final volume, \(V^{\prime}=\frac{M}{\rho^{\prime}}\)

Change in volume, \(V^{\prime}-V=M\left(\frac{\rho-\rho^{\prime}}{\rho^{\prime} \rho}\right)\)

∴ Volume strain = \(=\frac{V^{\prime}-V}{V}=\frac{\rho-\rho^{\prime}}{\rho^{\prime}}\)

∴ \(B=-\frac{p V}{V^{\prime}-V}=-\frac{p \times \rho^{\prime}}{\rho-\rho^{\prime}}\)

or, \(\underset{B}{p}=-\frac{\rho-\rho^{\prime}}{\rho^{\prime}}=\frac{\rho^{\prime}-\rho}{\rho^{\prime}}\)

or, \(\frac{\rho}{\rho^{\prime}}=1-\frac{p}{B}\)

∴ \(\frac{\rho^{\prime}}{\rho}=\frac{1}{1-\frac{p}{B}}\)

The option 2 is correct.

Elasticity Problems and Solutions for Class 11

Question 12. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl. on applying a force F, how much force is needed to stretch the second wire by the same amount?

1. 4F
2. 6F
3. 9F
4. F

Answer:

In case of first wire, Y = \(\frac{F / A}{\Delta l / l_0}=\frac{F l_0}{A \Delta l}\)

or, \(F=\frac{Y A \Delta l}{l_0}\)

In the case of the second wire,

Y = \(\frac{F^{\prime} / 3 A}{\frac{\Delta l}{l_0 / 3}}=\frac{F^{\prime} l_0}{9 A \Delta l}\)

or, \(F^{\prime}=\frac{9 Y A \Delta l}{l_0}=9 F\)

The option 3 is correct.

Question 13. Which type of substances are called elastomers? Give one example.
Answer:

Elastomers are those materials for which stress-strain variation is not a straight line within the elastic limit. An elastomer is a polymer with viscoelasticity (colloquially elasticity), generally having low Young’s modulus and high failure strain compared with other materials.

Example: Rubber.

Question 14. Bridges are declared unsafe after long use. Why?
Answer:

A bridge undergoes alternating stress and strain a large number of times during its use. A bridge loses its elastic strength when it is used for a long time. Therefore, the amount of strain for a given stress will become large and ultimately the bridge will collapse. So, they are declared unsafe after long use.

Question 15. What are elastomers? Give two examples for the same.
Answer:

Elastomers (elastic polymers) are materials of low Young’s modulus but of very high elastic limits. Such a material can withstand high strain but can still develop sufficient stress to bring it back to its initial size and shape.

Examples: natural rubber, thermoplastics.

Question 16. What is the value of rigidity modulus of elasticity for an incompressible liquid?
Answer:

A liquid, compressible or incompressible, does not have any defined shape; it cannot withstand shear. So it can never generate any shearing stress. Hence the rigidity modulus of elasticity of a liquid is zero.

Question 17. Which type of energy is stored in the spring of wrist wristwatch?
Answer:

Potential energy is stored in the spring of wrist watch.

Real-Life Examples of Elastic Materials

Question 18. The stress-strain graph for materials A and B are as shown in the graphs drawn to the same scale, which graph represents a property of ductile materials? Justify your answer.
Answer:

The graph for material A represents the property of ductile material because of its greater plastic range.

Question 19. Two wires A and B of length l, radius r, and length 21, radius 2 r having the same Young’s modulus Y are hung with a weight of mg as shown in the figure. What is the net elongation in the two wires?
Answer:

The length and radius of wire A are l and r and that of wire B are 2l and 2 r respectively.

If l₁ and l₂ are the individual elonga¬tion of wire A and wire B, then the net elongation,

∴ \(\Delta l =\Delta l_1+\Delta l_2=\frac{m g l}{\pi r^2 Y}+\frac{m g(2 l)}{\pi(2 r)^2 Y}\)

= \(\left(\frac{m g l}{\pi r^2 Y}+\frac{2 m g l}{4 \pi r^2 Y}\right)=\frac{4 m g l+2 m g l}{4 \pi r^2 Y}=\frac{3}{2} \frac{m g l}{\pi r^2 Y}\)

Question 20. Which of the two forces-deforming or restoring is responsible for the elastic behavior of a substance?
Answer: Restoring force is responsible for the elastic behavior of a substance.

First And Second Law Of Thermodynamics The Secondary Of Thermodynamics

The 2nd Law Of Thermodynamics

Let us consider a man holding a glass of hot tea. From our experience, we know that in such a case the hand gradually becomes hotter and the glass cooler. This is because some heat (say, 10 calories per second) is absorbed by the hand from the glass.

Second Law of Thermodynamics Explained

• Now, consider the opposite process. If 10 calories of heat is given by the cold hand to the hot glass per second, the energy would still be conserved, i.e., the first law of thermodynamics would still be obeyed. But, this opposite process never occurs in nature.
• In general, there is a natural direction in every real process. The first law of thermodynamics cannot determine this natural direction. So it is important to formulate a new law—the second law of thermodynamics.
• Scientists expressed the second law in different forms. However, all of the different forms are equivalent. They provide alternative statements of the same physical law.

Clausius and Kelvin Statements of the Second Law

Clausius’s statement: No self-acting machine can transfer heat from a lower to a higher temperature.

Kelvin-Flanck Statement: No self-acting machine can convert some amount of heat entirely into work.

Entropy: The glass and the hand together, is now regarded as a closed system. Energy can be transmitted from one part to another or can be transformed from one form to another in a closed system. But this type of process is irreversible. The direction of an irreversible process is determined by a change in a special property, called the entropy of the system.

The analysis of a reversible process, using the second law of thermodynamics, leads to the concept of entropy. Entropy, denoted by the letter S, is a property of all thermodynamic systems. It is defined by the relation

ds = \(\frac{dQ}{T}\)….(1)

where, dQ = heat exchange of a system in an infinitesimal reversible process at temperature T and dS = corresponding, change in entropy of the system.

From (1), dQ = TdS. Using this relation in the first law of thermodynamics, we get, dQ = dU+dW or, TdS = dU+pdV ….(2)

Now, we note that dQ and dW are quantities exchanged between the die system and the surroundings m a process, o thev depend on whether the process is reversible or irreversible.

But equation (2) does not contain any such exchange quarantine. So it is true for reversible as well as irreversible processes. Then, if we use equation (2) in thermodynamics, we need not worry about the nature of the process. This is the beauty of the ideal concept of reversibility.

The entropy principle: Thermodynamic analysis shows that in every real process in nature, the sum of the entropies of a system and its surroundings always increases. The opposite process, in which the sum of the entropies decreases, is not allowed in nature.

• We may compare the situation with the law of conservation of energy (first law of thermodynamics). This law states that the total energy of the universe is a constant—it can never increase or decrease. In analogy, the second law of.thermo¬dynamics states that the total entropy of the universe increases in every process it can never decrease.
• Every real process in nature occurs in such a direction that the total entropy of the universe increases. Alternatively, no process, in which the total entropy of the universe decreases, can occur in nature. This is known as the principle of increase of entropy.
• For a reversible process the total entropy of the universe remains constant while for an irreversible process. the entropy of the universe increases. As all natural processes in general are irreversible, every natural process results in an increase in entropy of the universe.

Entropy

This principle of increase of entropy is the most general statement of the second law of thermodynamics. The Clausius and the Kelvin-Planck statements can easily be derived from this principle.

In an adiabatic process, dQ = 0 . Then equation (1) gives that dS = 0 or, S = constant. So the entropy of a system remains constant in a reversible adiabatic process (just like temperature in an isothermal process). For this reason, a reversible adiabatic process is called an isentropic process.

In essence, each of the three laws of thermodynamics defines one important property of all thermodynamic systems:

1. Zeroth law: Temperature (T)
2. First law: Internal energy (U)
3. Second law: Entropy (S)

Unit 8 Thermodynamics Chapter 1 First And Second Law Of Thermodynamics Heat Engines And Refrigerators

Applications of the Second Law of Thermodynamics

Heat reservoir: A body whose temperature remains constant even when heat is gained or lost by it is called a heat reservoir. Every heat reservoir has its own characteristic temperature.

Generally, different heat reservoirs have different characteristic temperatures. They are drawn in the way as shown. The characteristic temperatures of the two reservoirs shown and T₁ and T₂ respectively.

• For example, the atmosphere or sea water may be taken as heat reservoir. If a burning oven is placed in the air, it rejects heat to the atmosphere. Or if a large block of ice is placed in the air, it receives heat from the atmosphere.
• From our daily experiences, it is known that we can ignore the change of temperature of the atmosphere in the above cases. Similarly, if a bucketful of boiling water is poured in the sea, sea water takes heat but its temperature does not change.
• It is obvious that the atmosphere and sea, being very large in size, behave as heat reservoirs. However, comparatively smaller bodies also may show similar properties.
• Suppose, the temperature of a coaloven when burning is 300°C. It continuously rejects heat to the surroundings. In spite of that, as long as coal burns, the temperature of the oven remains constant at 300°C. So in this case the oven acts as a heat reservoir.

We know, if dT is the change of temperature of a body due to a heat exchange of dQ, then the thermal capacity of the body, C = \(\frac{dQ}{dT}\).

In case of a heat reservoir, dT =0 so whatever the value of heat gain or heat loss (dQ) may be, C → ∞, i.e., the thermal capacity of any heat reservoir is infinite.

Conversely, it can be said that if the thermal capacity of a body is infinite, the body will behave as a heat reservoir.

Heat engine: It is a mechanical device that converts heat into work.

• In general, a heat engine H takes heat from a source at a higher temperature converts a part of it into work, and gives out the rest to a body (sink) at lower temperature.
• In most cases, the source at a higher temperature and the sink at a lower temperature are two heat reservoirs, i.e., in a complete cycle their temperatures are fixed at T₁ and T₂. Obviously, T₁>T₂.
• Generally, a heat engine works in a cyclic process. It means that, after the completion of a cycle by converting heat into work, the working substance of the engine returns to its initial condition and becomes ready for the next cycle.
• The action of each cycle is equivalent, i.e., in each cycle the same amount of heat is converted into the same amount of work.

Efficiency of a heat engine: The object of a heat engine is to convert heat taken from the source into useful work. So the heat taken from the source is called the input, and the transformed work is called the output.

The efficiency of a heat engine is defined as the fraction of total heat taken from the source which is converted into work. Suppose in each cycle,

the heat is taken by the engine from the source at temperature T₁ = Q_1;

transformed work = W;

heat rejected by the engine to its surroundings (sink) at temperature T₂ = Q₂.

From the principle of conservation of energy, we can write, Q₁ = W+Q₂ or, W = Q₁-Q₂

So, efficiency, \(\eta=\frac{\text { output }}{\text { input }}=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}\)

i.e., \(\eta=1-\frac{Q_2}{Q_1}\)…(1)

From equation (1) we get η ≤ 1, i.e., the efficiency of a heat engine can never be greater than 1 or 100%.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Second Law of Thermodynamics in Heat Engines

Ideal heat engine: it is easily understood that the more a heat engine converts heat into work the more its efficiency will be. If any engine converts the whole amount of heat taken from the source into work, then the engine is called an ideal heat engine.

It is evident that in case of this type of engine, the heat is rejected to its surroundings (Q₂)zero. So the work obtained will be equal to the heat taken from the source (Q₁). Therefore, the efficiency of an ideal heat engine,

⇒ \(\eta=\frac{W}{Q_1}=\frac{Q_1}{Q_1}=1=100 \%\)

Alternatively, \(\eta=1-\frac{Q_2}{Q_1}=1-\frac{0}{Q_1}=1=100 \%\)

Kelvin-Planck’s statement of the second law of thermodynamics: No self-acting machine can convert some amount of heat entirely into work. On the basis of the discussion about heat engines, this statement can be expressed in the form of an easy alternative: An ideal heat engine does not exist in nature.

Refrigerator: A mechanical device that transfers heat from a colder to a hotter place is called a refrigerator and the working substance of a refrigerator is called a refrigerant.

• Suppose, a cold body, say an icebox is at a temperature of T₂. The surrounding temperature T₁ in most cases is greater than T₂ So the temperature of the cold body begins to increase due to receiving heat from the surroundings.
• Now if any mechanical device removes heat at the same rate at which the cold body receives heat, the temperature of the body will remain constant i.e., the body will remain in the same cold state. This mechanical arrangement is called a refrigerator, denoted by R in.
• Generally, with the help of some external work, the refrigerator transfers heat from the lower temperature T₂ to the higher temperature T₁ of the surroundings. Our household refrigerator is a familial example of this machine.

Like a heat engine, a refrigerator also works in a cyclic process. At the end of each cycle, the working substance returns to its initial state, and then the next cycle starts.

Coefficient of performance of a refrigerator: The aim of a refrigerator is to extract heat from a cold body at the expense of some external mechanical work. The external supplied work is known as the input and the heat extracted from the cold body is called the output.

Suppose, in each complete cycle, heat received by the refrigerator from the colder body at temperature T₂ = Q₂;

heat delivered by the refrigerator to the surroundings at higher temperatures T₁ = Q₁;

external work done = W

From the principle of conservation of energy, we can write,

W + Q₂ = Q₁

or, W = Q₁-Q₂

From the definition of the coefficient of performance (e) of a refrigerator, we have,

∴ e = \(\frac{\text { output }}{\text { input }}=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}\)….(2)

Ideal refrigerator: Obviously, the less the amount of external work supplied to a refrigerator to run it, the better is its performance. If any refrigerator can transfer heat from low temperature to a higher temperature without any help of external work, then it is called an ideal refrigerator. For example, if any household refrigerator could run without any assistance of electricity, then that would be an ideal refrigerator.

For this ideal refrigerator, W = 0, i.e., Q₁ – Q₂ = 0 or, Q₁ = Q₂

So, coeffieicient of performance of an ideal refrigerator, e = \(\frac{Q_2}{W}=\frac{Q_2}{0} \rightarrow \infty\)

i.e., the coefficient of performance of an ideal refrigerator is infinite.

Clausius’s statement of the second law of thermodynamics: No self-acting machine can transfer heat from a lower to a higher temperature. On the basis of the discussions about refrigerators, this statement can be expressed in the form of an easy alternative: An ideal refrigerator does not exist in nature.

Refrigerator Discussions:

1. Comparing It is apparent that the working principles of a heat engine and a rein aerator are exactly opposite to each other. But it does not mean that these two machines are used for opposite purposes.
   • A heat engine is a machine for conversion of heat into work, but a refrigerator is not a machine for conversion of work into heat.
   • On the omer nana. I refrigerator is a machine that transfers heat from a low temperature to a higher temperature, but heat engines are never used to transfer heat from a higher to a lower temperature.
2. Again, on comparing it is found that if it is possible to conduct the different operations of a heat engine (ie.. heat intake from higher temperature.  transformation of heat into work, and rejection or heat at low temperature) in the reverse direction, then it will act as a refrigerator.
   • The necessary condition for this is that each operation should have to be a reversible process. Le.. the engine should have to be a reversible heat engine. In that case, the refrigerator running in the opposite direction will be reversible.
3. A reversible process is nothing but an ideal process. In nature this process does not exist—all real processes are irreversible. So, there is no opportunity to use a heat engine as a refrigerator, and vice versa.

First And Second Law Of Thermodynamics Useful Relations For Using Solving Problems

If W = work done and H = corresponding heat produced, then from Joule’s law, W ∝ H or, W = JH

where J = constant = mechanical equivalent of heat = 4.2 J · cal^-1 = 4.2 x 10⁷ erg  · cal^-1.

1 cal = 4.2 J = 4.2 x 10⁷ erg.

According to the first law of thermodynamics, heat taken by the system from the surroundings = change in internal energy + external work done

Q = (U_f – U_i) + W (integral form)

dQ = dU+ dW (differential form)

Infinitesimal work done by a system : dW = pdV

Work done in a finite process, W = ∫dW = ∫pdf

• c_v = specific heat of a substance at constant volume. The heat taken in a process is Q = \(m c_v t\), where m = mass of the substance and t = increase in temperature when the volume remains constant.
• Molar specific heat at constant volume is Cv = \(m c_v\), where M = molecular weight of the substance.
• c_p = specific heat of a substance at constant pressure. The heat taken in a process is Q = mc_pt, where m = mass of the substance and t = increase in temperature when the pressure remains constant.
• Molar specific heat at constant pressure is \(C_p=M c_p\), where M molecular weight of the substance.
• For an ideal gas, the difference between the molar specific heat is C_p – C_v = R
• If 1g is taken instead of 1 mol, then \(c_p-c_v=\frac{R}{M}\) m = molecular weight.

The ratio between the two specific heats is \(\gamma=\frac{C_p}{C_y}\) Also, C_p > C_v, γ> 1

For an isothermal process of n mol of an ideal gas, Q = \(W=n R T \ln \frac{V_f}{V_i}=n R T \ln \frac{p_i}{p_f}\)

In an adiabatic process of an ideal gas, p, V, and T are related as \(p V^\gamma\)=constant, \(T V^{\gamma-1}\)=constant, \(T^\gamma p^{1-\gamma}=\text { constant } \text {. }\) = constant.

Adiabatic work done for n mol of an ideal gas is W = \(n C_v\left(T_i-T_f\right)=\frac{n R}{\gamma-1}\left(T_i-T_f\right)=\frac{p_i V_i-p_f V_f}{\gamma-1}\)

Efficiency of a heat engine, \(\eta=1-\frac{Q_2}{Q_1}\)

where Q₁ =heat taken by the engine from the source at temperature T₁

and Q₂ =heat rejected by the engine to its surroundings (sink) at temperature T₂

Coefficient of performance of a refrigerator, e = \(\frac{Q_2}{Q_1-Q_2}\)

where Q₂ = heat received by the refrigerator from the colder body at temperature T₂,

and Q₁ = heat delivered by the refrigerator to the surroundings at higher temperatures T₁

The efficiency of a Carnot engine using an ideal gas. \(\eta=1-\frac{T_2}{T_1}\)

where T₁ = temperature of the source and T₂ = temperature of the sink

In the case of a Carnot refrigerator, work done and heat exchange are equal and opposite to the corresponding quantities of a Carnot engine. So, in this case, the coefficient of performance of a refrigerator,

∴ e = \(\frac{T_2}{T_1-T_2}\)

First And Second Law Of Thermodynamics Long Answer Type Questions

Long Answer Questions on First and Second Laws of Thermodynamics for Class 11

Question 1. Water falls from the top to the bottom of a waterfall. Why does the temperature at the bottom become slightly higher?
Answer:

Given

Water falls from the top to the bottom of a waterfall.

Water at the top has a potential energy due to its height. During free fall, the potential energy is converted into kinetic energy. On impact with the ground the kinetic energy of water is converted mainly into heat energy. The heat evolved increases the temperature of the water slightly.

Question 2. A bullet becomes hot on hitting a target. Why?
Answer:

Given

A bullet becomes hot on hitting a target.

A bullet is obstructed when it hits a target. As a result of the impact, the kinetic energy of the bullet is converted mainly into heat. This heat evolved increases the temperature of the bullet, as well as that of the target.

Read and Learn More Class 11 Physics Long Answer Questions

Question 3. Two balls of the same mass, one of iron and the other of copper, are dropped from the same height. Which one would become hotter?
Answer:

Given

Two balls of the same mass, one of iron and the other of copper, are dropped from the same height.

As the mass and the height of both balls are the same, the initial potential energy (= mgh) is the same for both balls. So the kinetic energy on impact with the ground will also be the same. Then, the same amount of heat will be produced.

We know that, increase in temperature = \(\frac{\text { heat produced }}{\text { mass } \times \text { specific heat }}\)

However, the specific heat of copper is lower than that of iron. So the copper ball will be hotter due to its higher rise in temperature.

Question 4. Show that the external work done by a gas in an isothermal expansion is equal to the heat supplied to the gas.
Answer:

For an ideal gas, the internal energy depends only on its temperature. For a real gas, the change in internal energy at constant temperature is negligible. So for an isothermal expansion (T = constant), U_f = U_i. Then, from the first law of thermodynamics,

U_f– U_i = Q-W or, 0 = Q-W or, W= Q

This means that the work done by the gas is equal to the heat supplied to it.

Question 5. ‘An isothermal process is essentially a very slow process’. Explain.
Answer:

‘An isothermal process is essentially a very slow process’.

The volume and the pressure of a system change in an isothermal process, but the temperature remains constant.

• For example, during the expansion of a gas, as the internal energy of the gas is converted into work, the temperature of the gas tends to fall.
• But if the expansion is very slow, the gas gets sufficient time to take heat from the surroundings. This heat does the work and the internal energy is not used up. Hence, the temperature remains constant.
• Similarly, during compression, work is done on the gas. Hence, heat is generated in the gas. This heat increases the temperature of the gas.
• But in a very slow compression, the gas gets sufficient time to lose this heat to the surroundings. As a result, the temperature remains constant.
• So, an isothermal process is essentially a very slow process.

Second Law of Thermodynamics: Long Answer Format

Question 6. What is the source of the energy that does the work in an adiabatic expansion of a gas?
Answer:

From the first law of thermodynamics, U_f – U_i = Q-W, i.e., change of internal energy = heat gained – work done.

For an adiabatic expansion, Q= 0.

So, U_f -U_i = -W or, W = U_i – U_f

i.e., work done = decrease in internal energy. This means that internal energy is converted into external work in an adiabatic expansion and the temperature of the gas decreases.

Question 7. A gas is compressed to half its volume in two different ways:

1. Very rapidly and
2. Very slowly. In which process will the work done on the gas be higher?

Answer:

Let the initial state of the gas be represented by point A on a pV diagram.

1. When the gas is compressed very rapidly, it is an adiabatic process. This compression from volume V to volume \(\frac{V}{2}\) is represented by the curve AB. So the work done on the gas – area under the curve AB = area ABDFA.

2. When the gas is compressed very’ slowly, it is an isothermal process. This compression is represented by the curve AC. So the work done = area ACDEA.

The adiabatic curve is always steeper than the isothermal curve at a point. So line AB is above the line AC in the diagram. Then clearly, area ABDEA > area ACDEA. This means that a higher amount of work will be done on the gas in the rapid process.

Question 8. A gas is compressed to half its volume

1. Adiabatically,
2. Isothermally. In which process will the final temperature be higher?

Answer:

The temperature does not change in an isothermal process.

1. In adiabatic compression, work is done on the gas. This work increases its internal energy. As a result, the temperature of the gas also increases.
2. So, the final temperature would be higher for adiabatic compression.

Examples of Long Answer Questions on Thermodynamic Laws

Question 9. A gas expands from state 1 to another state 2 of higher pressure in two different processes as shown. Which process would require a greater supply of heat from the surroundings?

Answer:

Work done in each constant volume part of the two processes is zero.

For the first process, the work done in the constant pressure part = p₂ (V₂ -V₁).

From the first law of thermodynamics, the heat supplied from the surroundings during the first process, \(Q_1=\left(U_2-U_1\right)+W=\left(U_2-U_1\right)+p_2\left(V_2-V_1\right)\)

Similarly, the heat supplied from the surroundings in the second process, \(Q_2=\left(U_2-U_1\right)+p_1\left(V_2-V_1\right)\)

Now, (U₂ – U₁) is the same for the two processes as the initial and file final states are the same. As the final state is at a higher pressure, we have \(p_2>p_1 \text {. So, } Q_1>Q_2 \text {. }\).

So the first process would take a greater amount of heat from the surroundings.

Example 10. A screen divides a container of volume 1 m³ into two parts. One part is filled with an ideal gas at 300 K and the other part is empty. The system is isolated from the surroundings. Will there be any change in the temperature of the gas if the screen is suddenly removed?
Answer:

Given

A screen divides a container of volume 1 m³ into two parts. One part is filled with an ideal gas at 300 K and the other part is empty. The system is isolated from the surroundings.

The system is isolated; so heat exchange with the surroundings, Q = 0.

There is no change in the volume of the container; so work done, W = 0

From the first law of thermodynamics,

U_f – U_i = Q- W= 0 or, U_f = U_i

This means that the internal energy of the gas does not change. For an ideal gas, the internal energy depends only on its temperature. So the temperature of the gas remains unchanged at 300 K.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 11. For an adiabatic process of an ideal monatomic gas, the pressure and the temperature are related as p ∝ T^c. Find out the value of C.
Answer:

Given

For an adiabatic process of an ideal monatomic gas, the pressure and the temperature are related as p ∝ T^c.

For an adiabatic process of an ideal gas, \(p^{1-\gamma} T^\gamma= constant =k\) (say).

or, \(p^{1-\gamma}=k T^{-\gamma} or, \quad p=\left(k T^{-\gamma}\right)^{\frac{1}{1-\gamma}}=k^{\frac{1}{1-\gamma}} T^{\frac{\gamma}{\gamma-1}}\)

∴ \(p \propto T^{\frac{\gamma}{\gamma-1}}\)

As \(p \propto T^C, C=\frac{\gamma}{\gamma-1}\).

For a monatomic gas, \(\gamma=\frac{5}{3}\).

So, C = \(\frac{\frac{5}{3}}{\frac{5}{3}-1}=\frac{5}{2}\).

Question 12. Is the solar system in thermal equilibrium?
Answer:

No, the solar system is not in thermal equilibrium. The sun continuously radiates heat energy in all directions. Every planet, satellite, or object in the solar system absorbs this radiated heat. If the solar system is in thermal equilibrium, no heat flow will occur. This will cause the thermal death of the solar system.

Question 13. The pressure and the temperature of an ideal gas in an adiabatic process are related as p ∝ T³. What is the value of the ratio  C_P/C_v of the gas?
Answer:

Given

The pressure and the temperature of an ideal gas in an adiabatic process are related as p ∝ T³.

For 1 mol of an ideal gas,

pV= RT or, T = \(\frac{pV}{R}\)

Now, \(p \propto T^3\)

or, p = \(kT^3=k\left(\frac{p V}{R}\right)^3\)[k= constant]

∴ p = \(k \frac{p^3 V^3}{R^3} or, p^2 V^3=\frac{R^3}{k}\)

or, \(p V^{3 / 2}=\left(\frac{R^3}{k}\right)^{1 / 2}\)

or, \(p V^{3 / 2}\)= constant

For an adiabatic process, \(p V^\gamma=\) constant

So, \(\gamma=\frac{C p}{C}=\frac{3}{2}\).

Question 14. At constant pressure, the temperature coefficient of volume expansion of an ideal gas is \(\delta=\frac{1}{V} \frac{d V}{d T}.\). What will be the nature of the graph relating δ with the temperature T?
Answer:

Given

At constant pressure, the temperature coefficient of volume expansion of an ideal gas is \(\delta=\frac{1}{V} \frac{d V}{d T}.\).

For 1 mol of an ideal gas,

pV = \(R T \text { or, } V=\frac{R T}{p}\).

So at constant pressure, \(\frac{d V}{d T} =\frac{R}{p}\)

∴ \(\delta =\frac{1}{V} \frac{d V}{d T}=\frac{1}{V} \frac{R}{p}=\frac{R}{p V}\)

= \(\frac{R}{R T}=\frac{1}{T}\)

or, δ T= constant

This equation is represented by a rectangular hyperbola on the δ-T graph.

Calculating Work and Heat Transfer: Long Answers

Question 15. A rapid compression heats a gas, but a rapid expansion cools it—why?
Answer:

Given

A rapid compression heats a gas, but a rapid expansion cools it

A rapid thermal process can be regarded as an adiabatic process because the gas does not get sufficient time to exchange heat with the surroundings.

So, Q = 0.

From the first law of thermodynamics, U_f – U_i = Q-W = 0 -W = -W.

∴ W= U_i – U_f

1. In adiabatic compression, W is negative. So \(U_i-U_f<0, \quad \text { or, } \quad U_i<U_f\). This means that the internal energy of the gas increases. The internal energy of a gas depends only on its temperature. So the temperature increases and the gas is heated.

2. Conversely, in an adiabatic expansion, W is positive. So, U_f < U_i i.e., the internal energy decreases. As a result, temperature decreases; so the gas is cooled.

Question 16. Why does a bicycle pump become hot when it pumps air in a bicycle tube?
Answer:

The pumping operation is very rapid. So, it is essentially an adiabatic compression. Then, Q = 0, and W is negative.

From the first law of thermodynamics, \(U_f-U_i=Q-W=0-W \text { or, } W=U_i-U_f\)

As W is negative, \(U_i-U_f<0 or, U_i<U_f\).

So, the internal energy of air increases.

As a result, the temperature also increases and the pump becomes hot.

Question 17. Adiabatic and isothermal processes of an ideal gas are represented on a pV diagram by two curves, A and B. Can we say that curve A represents the isothermal process?

Answer:

Given

Adiabatic and isothermal processes of an ideal gas are represented on a pV diagram by two curves, A and B.

The statement is correct. We know that at every point on a pV diagram, the adiabatic curve is steeper than the isothermal curve. Curve B is steeper than curve A. So B represents the adiabatic process and A represents the isothermal process.

Question 18. In the V-T diagram, two different points show pressures p₁ and p₂. Which pressure would be higher p₁ or p₂? 

Answer:

In the V-T diagram, two different points show pressures p₁ and p₂.

The V- T graph is a straight line.

Let the equation of the straight line be V = aT+ b.

For 1 mol of an ideal gas, pV = RT.

So, \(p=\frac{R T}{V}=\frac{R T}{a T+b}=\frac{R}{a+\frac{b}{T}}\)

According to this relation, p increases when T increases.

T₂ > T₁.

So, p₂ > p₁.

Key Concepts in Thermodynamics: Long Answer Questions

Question 19. The volume of a gas is doubled

1. Very rapidly, and
2. Very slowly. In which one of the processes is the work done by gas higher?

Answer:

The rapid process is adiabatic and the slow process is isothermal. As the adiabatic curve is steeper than the isothermal curve, AB and AC represent the isothermal and adiabatic processes, respectively.

• Now, work done is given by the area under the curve corresponding to a process.
• As area ABDE > area ACDE, the work done in the isothermal process is greater than that in the adiabatic process.

Question 20. How will the temperature of an ideal gas change in an adiabatic expansion?
Answer:

In an adiabatic process, Q = 0

For the expansion of a gas, W is positive.

∴ Q = (U_f – U_i)+ W

or, 0 = (U_f – U_i) + W

or, U_f – U_i = -W

So, U_f< U_i, i.e., internal energy decreases. As the internal energy of a gas depends only on its temperature, the temperature decreases.

WBCHSE Class 11 Physics For First And Second Law Of Thermodynamics Short Answer  Type Questions

Short Answer Questions on First and Second Laws of Thermodynamics for Class 11

Question 1. An ideal gas undergoes a cyclic process (V, P) → (V, P) → (V, 4P) → (V, P) along straight lines in the P- V plane. Calculate the work done in the process.
Answer:

Given

An ideal gas undergoes a cyclic process (V, P) → (V, P) → (V, 4P) → (V, P) along straight lines in the P- V plane.

Area of the cycle = 1/2 x base x height

= 1/2(3V -V)(4P-P) = 3PV

Ad the cycle is anticlockwise, the work done = -3PV

Question 2. The temperature of a fixed mass of ideal gas is changed from T₁K to T₂K (T₂ > T₁). Calculate the work done if the change is brought about at

1. Constant pressure,
2. Constant volume.

Hence find from the first law of thermodynamics, in which case heat absorbed will be greater.

Answer:

1st law of thermodynamics: Q = (U₂-U₁)+ W

Work done at constant volume, W₁ = 0, Q₁ = C_v(T₂-T₁)

Change in internal energy = U₂ – U₁ = C_v(T₂ -T₁)

At constant pressure, Q₂ = C_p(T₂ – T₁)

For the same change in temperature, there will be some change in internal energy.

So the work done in this case, \(W_2=Q_2-\left(U_2-U_1\right)=C_p\left(T_2-T_1\right)-C_v\left(T_2-T_1\right)\)

= \(\left(C_p-C_v\right)\left(T_2-T_1\right)\)

As, \(C_p>C_v, W_2>0\). So, in this case \(Q_2=W_2+C_v\left(T_2-T_1\right)\)

∴ \(Q_2>Q_1\)

So in case of constant pressure heat absorbed will be greater.

Question 3. 1 mol of an ideal gas is compressed to half of its initial volume

1. Isothermallv and
2. Adiabatically.

Draw the p-V diagram in each case and hence state with reason, in which case the work done is less.

Answer:

The slope of the adiabatic curve is more than the isothermal curve.

Work done in the isothermal process = -area of ABCD and

the work done in the adiabatic process =- area of ABCU.

Comparing the two areas the work done in isothermal compression will be less.

Question 4. The second law of thermodynamics

1. Gives the definition of temperature.
2. Determines the direction of flow of heat during heat exchange between bodies.
3. Is another form of die principle of conservation of heat and other forms of energy.
4. Helps in computing the efficiency of the Camot’s engine.

Answer:

The options 1, 2, and 4 are correct.

Understanding First Law of Thermodynamics: Short Answers

Question 5. A certain amount of an ideal gas (γ= 1.4) performs 80 I of work while undergoing isobaric expansion. Find the amount of heat absorbed by the gas in the process and the change in its internal energy.
Answer:

Given

A certain amount of an ideal gas (γ= 1.4) performs 80 I of work while undergoing isobaric expansion.

For nmol of a gas,

pV = nRT or, pdV = nRdT [p constant]

or, \(d W=n R d T \quad or, d T=\frac{d W}{n R}\)….(1)

Also, \(\frac{C_p}{C_v}=\gamma\) or, \(\frac{C_p-C_v}{C_v}=\frac{\gamma-1}{1}\)

or, \(C_v=\frac{n R}{\gamma-1} \quad\left[because C_p-C_v=n R\right]\)

We know, change in internal energy, \(d U=C_v d T\)

From (1), (2) and (3)

dU = \(\frac{n R}{\gamma-1} \times \frac{d W}{n R}=\frac{80}{1-1.4}=200 \mathrm{~J}\)

Question 6. The efficiency of a Carnot engine is 50%. The temperature of the heat sink is 27 C.  Find the temperature of the heat source.
Answer:

Given

The efficiency of a Carnot engine is 50%. The temperature of the heat sink is 27 C.

The temperature of the heat sink

T₂ = 27 + 273 = 300K

Efficiency, \(\eta=50 \%=\frac{50}{100}=\frac{1}{2}\)

If the temperature of the heat source is T, \(\eta=1-\frac{T_2}{T_1}\)

i.e., \(T_1=\frac{T_2}{1-\eta}=\frac{300}{1-\frac{1}{2}}\)

= 600K = (600 – 273) °C =327 °C

Second Law of Thermodynamics: Short Answer Format

Question 7. During adiabatic expansion of 2 mol of a gas, the internal energy of the gas is found to decrease by 2J. The work done by the gas during the process is

1. 1J
2. -1J
3. 2J
4. -2J

Answer:

Given

During adiabatic expansion of 2 mol of a gas, the internal energy of the gas is found to decrease by 2J.

Work done in adiabatic process = decrease in internal energy

= 2J

The option 3 is correct.

Question 8. The slope of an isothermal curve is always

1. The same as that of an adiabatic curve
2. Greater than that of an adiabatic curve
3. Less than that of an adiabatic curve
4. None of these

Answer: 3. Less than that of an adiabatic curve

Then option 3 is correct.

Question 9. When 110 J of heat is supplied to a gaseous system, the internal energy of the system increases by 40 J. The amount of external work done (in f) is

1. 150
2. 70
3. 110
4. 40

Answer:

Given

When 110 J of heat is supplied to a gaseous system, the internal energy of the system increases by 40 J.

According to the first law of thermodynamics,

dQ = dU+ dW

or, dW= dQ-dU = 110-40 = 70

The option 2 is correct.

Examples of Short Answer Questions on Thermodynamic Laws

Question 10. One mole of an ideal monatomic gas is heated at a constant pressure from 0 °C to 100 °C. Then the change in the internal energy of the gas is (given R = 8.32 J · mol · k^-1)

1. 0.83 x 10³ J
2. 4.6 x 10³ J
3. 2.08 x 10³ J
4. 1.25 x 10³ J

Answer:

Given

One mole of an ideal monatomic gas is heated at a constant pressure from 0 °C to 100 °C.

Change in internal energy, \(\Delta U =n C_v\left(T_f-T_i\right)=1 \times \frac{3}{2} R \times 100\)

= \(1 \times \frac{3}{2} \times 8.32 \times 100=1.25 \times 10^3 \mathrm{~J}\)

The option 4 is correct.

Question 11. One mole of a van der Waals gas obeying the equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\), undergoes the quasistatic cyclic process which is shown in the p- V diagram. The net heat absorbed by the gas in this process is

1. \(\frac{1}{2}\left(p_1-p_2\right)\left(V_1-V_2\right)\)
2. \(\frac{1}{2}\left(p_1+p_2\right)\left(V_1-V_2\right)\)
3. \(\frac{1}{2}\left(p_1+\frac{a}{V_1^2}-p_2-\frac{a}{V_2^2}\right)\left(V_1-V_2\right)\)
4. \(\frac{1}{2}\left(p_1+\frac{a}{V_1^2}+p_2+\frac{a}{V_2^2}\right)\left(V_1-V_2\right)\)

Answer:

Given

One mole of a van der Waals gas obeying the equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\), undergoes the quasistatic cyclic process which is shown in the p- V diagram.

In the case of a cyclic process, the total heat absorbed by the gas

= work done in the total cycle

= area of the cycle

= \(\frac{1}{2}\left(p_1-p_2\right)\left(V_1-V_2\right)\)

The option 1 is correct.

Calculating Work and Heat Transfer: Short Answers

Question 12. A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in the diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element such that the temperature rises with time t as \(\Delta T=\alpha t+\frac{1}{2}\)βt² (α and β are constant), while the pressure remains constant. The atmospheric pressure above the piston is p₀, Then

1. The rate of increase in internal energy is \(\frac{5}{2} r(\alpha+\beta t)\)
2. The current flowing in the element is \(\sqrt{\frac{5}{2 r} r(\alpha+\beta t)}\)
3. The piston moves upwards with constant acceleration
4. The piston moves upwards at a constant speed

Answer:

Given

A heating element of resistance r is fitted inside an adiabatic cylinder which carries a frictionless piston of mass m and cross-section A as shown in the diagram. The cylinder contains one mole of an ideal diatomic gas. The current flows through the element such that the temperature rises with time t as \(\Delta T=\alpha t+\frac{1}{2}\)βt² (α and β are constant), while the pressure remains constant.

Increase in internal energy, \(\Delta U =n C_v \Delta T=1 \times \frac{5}{2} R \times\left(\alpha t+\frac{1}{2} \beta t^2\right)\)

= \(\frac{5}{2} R\left(\alpha t+\frac{1}{2} \beta t^2\right)\)

Rate of increase in internal energy, \(\frac{d(\Delta U)}{d t}=\frac{5}{2} R(\alpha+\beta t)\)

According to Charles’ law, \(\Delta V \propto \Delta T\) [when pressure constant]

∴ \(\Delta V=C\left(\alpha t+\frac{1}{2} \beta t^2\right)\) [C=constant]

So, displacement of the piston,

∴ \(\Delta x=\frac{\Delta V}{A}=C^{\prime}\left(\alpha t+\frac{1}{2} \beta t^2\right) \quad\left[C^{\prime}=\frac{C}{A}=\text { constant }\right]\)

Hence, the displacement of the piston obeys the equation of motion of a particle in constant acceleration.

The options 1 and 3 are correct.

Question 13. The pressure p, volume V, and temperature T for a certain gas are related \(p=\frac{A T-B T^2}{V}\), where A and B are constants. The work done by the gas when the temperature changes from T₁ to T₂ while the pressure remains constant is given by

1. \(A\left(T_2-T_1\right)+B\left(T_2^2-T_1^2\right)\)
2. \(\frac{A\left(T_2-T_1\right)}{V_2-V_1}-\frac{B\left(T_2^2-T_1^2\right)}{V_2-V_1}\)
3. \(A\left(T_2-T_1\right)-B\left(T_2^2-T_1^2\right)\)
4. \(\frac{A\left(T_2-T_2^2\right)}{V_2-V_1}\)

Answer:

Given

The pressure p, volume V, and temperature T for a certain gas are related \(p=\frac{A T-B T^2}{V}\), where A and B are constants.

Work done, \(W \int p d V=\int_{V_1}^{V_2} d V=p\left(V_2-V_1\right)\)

According to the given relation, \(V_1 =\frac{A T_1-B T_1^2}{p}, V_2=\frac{A T_2-B T_2^2}{p}\)

∴ W = \(p\left(\frac{A T_2-B T_2^2}{p}-\frac{A T_1-B T_1^2}{p}\right)\)

= \(A\left(T_2-T_1\right)-B\left(T_2^2-T_1^2\right)\)

The option 3 is correct.

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 14. 2 mol of an ideal monatomic gas is carried from a state (P₀, V₀) to a state (2P₀, 2V₀) along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

1. \(3 P_0 V_0\)
2. \(\frac{9}{2} P_0 V_0\)
3. \(6 P_0 V_0\)
4. \(\frac{3}{2} P_0 V_0\)

Answer:

Given

2 mol of an ideal monatomic gas is carried from a state (P₀, V₀) to a state (2P₀, 2V₀) along a straight line path in a P-V diagram.

According to the graph, if temperatures of state A and state B of the gas are T₁ and T₂, then \(P_0 V_0=n R T_1 \text { and } 2 P_0 \times 2 V_0=n R T_2\)

∴ \(T_2-T_1=\frac{4 P_0 V_0}{n R}-\frac{P_0 V_0}{n R}=\frac{3 P_0 V_0}{n R}\)

From state A to B at the intermediate state, the pressure is changed from P₀ to 2P₀ by changing the temperature from T₁ to T₂ keeping the volume constant.

Then increase in internal energy of the gas,

ΔU = the amount of heat absorbed by n mol gas at constant volume

= n x C_v x(T₂-T₁)

= \(n \times \frac{3 R}{2} \times \frac{3 P_0 V_0}{n R}\) [for monatomic gas, \(C_\nu=\frac{3 R}{2}\)

= \(\frac{9 P_0 V_0}{2}\)

Work done by the gas absorbing the remaining heat, ΔW= area of ABDC

= \(\left(2 V_0+V_0\right) \frac{2 P_0-P_0}{2}=\frac{3 P_0 V_0}{2}\)

∴ \(\Delta Q =\Delta U+\Delta W\)

= \(\frac{9}{2} P_0 V_0+\frac{3}{2} P_0 V_0=6 P_0 V_0\)

The option 3 is correct.

Question 15. One mole of a monatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points (V₀, T₀) and (2V₀, 3T₀) in a V-T diagram. What is the value of the heat capacity of the gas at the point (V₀, T₀)?

1. R
2. 3/2 R
3. 2R
4. 0

Answer:

Given

One mole of a monatomic ideal gas undergoes a quasistatic process, which is depicted by a straight line joining points (V₀, T₀) and (2V₀, 3T₀) in a V-T diagram.

dQ = dU + pdV

or, \(d Q=n C_v d T+p d V\)

or, \(d Q=C_v d T+p d V\) [given n=1mol] ….(1)

For monatomic gas, \(\gamma=\frac{5}{3} \quad \text { or, } \frac{C_p}{C_v}=\frac{5}{3} \quad \text { or, } C_p=\frac{5}{3} C_v\)

Now, \(C_p-C_v=R\)

or, \(\frac{5}{3} C_v-C_v=R or, C_v=\frac{3}{2} R\)

∴ From (1) we get,

dQ = \(\frac{3}{2} R \times\left(3 T_0-T_0\right)+\frac{R T_0}{V_0} \times\left(2 V_0-V_0\right)\)

or, \(d Q=4 R T_0\)

or, \(C d T=4 R T_0\)[C is heat capacity]

or, \(C \times 2 T_0=4 R T_0\) or, C=2 R

The option 3 is correct.

Question 16. For an ideal gas with initial pressure and volume P_i and V_i respectively, a reversible isothermal expansion happens, when its volume becomes V₀. Then it is compressed to its original volume V_i by a reversible adiabatic process. If the final pressure is P_f then which of the following statement is true?

1. \(P_f=P_i\)
2. \(P_f>P_i\)
3. \(P_f<P_i\)
4. \(\frac{P_f}{V_0}=\frac{P_i}{V_i}\)

Answer:

Given

For an ideal gas with initial pressure and volume P_i and V_i respectively, a reversible isothermal expansion happens, when its volume becomes V₀. Then it is compressed to its original volume V_i by a reversible adiabatic process.

We know, the slope of the adiabatic curve > the slope of the isothermal curve.

When the gas of volume V₀ is compressed adiabatically to its initial volume V_i, the slope of the curve will be greater than an isothermal curve.

Here final pressure P_f at initial volume V_i will be greater than P_i.

∴ P_f > P_i

The option 2 is correct.

Practice Short Answer Questions on Thermodynamics for Class 11

Question 17. Which of the following statement(s) is/are true? “Internal energy of an ideal gas ______”

1. Decreases in an isothermal process
2. Remains constant in an isothermal process
3. Increases in an isobaric process
4. Decreases in an isobaric expansion

Answer:

The internal energy of an ideal gas depends only on the temperature of the gas.

The option 2 is correct.

Question 18. One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown. The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800K, and 600K, respectively, Choose the correct answer

1. The change in internal energy in the whole cyclic process is 250 R
2. The change in internal energy in the process CA is 700R
3. The change in internal energy in the process AB is -350R
4. The change in internal energy in the process BC is -500R

Answer:

Given

One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown. The process BC is adiabatic. The temperatures at A, B, and C are 400 K, 800K, and 600K, respectively,

Change in internal energy in

1. Process CA is \(U_A-U_C =C_\nu(400-600)=-200 C_v\)

= \(-200 \times \frac{5}{2} R\)

=-500 R

2. Process AB is \(U_B-U_A =C_\nu(800-400)=400 \times \frac{5}{2} R\)

= 1000 R

3. process BC is \(U_C-U_B=C_v(600-800)=-200 \times \frac{5}{2} R\)

=-500 R

4. The cycle ABCA is \(U_A-U_A=0\).

We used the property of an ideal gas that U is a function of temperature only.

Then from the 1st law of the hemodynamics, Q= (U₂ -U₁)+ W

We get for a process at constant volume, \(C_v\left(T_2-T_1\right)=\left(U_2-U_1\right)+0\)

i.e., \(U_2-U_1=C_v\left(T_2-T_1\right)\)

For a diatomic ideal gas, \(C_\nu=\frac{5}{2} R\)

The option 4 is correct.

Question 19. A solid body of constant heat capability of 1 J/°C, Is being heated by keeping it In contact with reservoirs in two ways:

1. Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies the same amount of beat.
2. Sequentially keeping In contact with 11 reservoirs such that each reservoir supplies the same amount of beat.

In both cases, body is brought from the initial temperature of 100°C to the final temperature of 200°C. Entropy change of the body in the two cases respectively Is

1. In 2, 4ln 2
2. In 2, In 2
3. In 2, 2ln 2
4. 21n 2, Bln 2

Answer:

100°C = 373K, 200°C = 473K

Increase in temperature 100°C= 100K

Heat capacity, C = 1 J/K

1. For each heat reservoir,

increase in temperature = 100/2 = 50 K

So, entropy change,

⇒ \(\Delta S_1=C\left[\int_{373}^{423} \frac{d T}{T}+\int_{423}^{473} \frac{d T}{T}\right]\)

= \(\ln \frac{423}{373}+\ln \frac{473}{423}\)

= \(\ln \frac{473}{373}\)

2. For each heat reservoir,

Increase in temperature = 100/8 = 12.5 K

So, entropy change,

⇒ \(\Delta S_2=C\left[\int_{373}^{385.5} \frac{d T}{T}+\cdots+\int_{460.5}^{473} \frac{d T}{T}\right]\)

= \(\ln \frac{473}{373}\)

(Note: The values of ΔS₁, and ΔS₂ do not match any of the given answers. If we take the temperatures to be 100 K and 200 K instead of 100°C and 200°C, then ΔS₁ = ΔS₂ = In2)

The option 2 is correct.

Question 20. An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. During tills process the relation of pressure p and volume V is given by pVⁿ – constant, then n is given by (Here C_p and C_v are molar sped He heat at constant pressure and constant volume, respectively)

1. \(n-\frac{C_p}{C_v}\)
2. \(n=\frac{C \cdot C_p}{C \cdot C_\nu}\)
3. \(n=\frac{C_p C}{C-C_v}\)
4. \(n-\frac{C-C_y}{C C_p}\)

Answer:

Given

An ideal gas undergoes a quasistatic, reversible process in which its molar heat capacity C remains constant. During tills process the relation of pressure p and volume V is given by pVⁿ – constant,

Here, pVⁿ = k (constant)….(1)

For 1 mol of an ideal gas, pV = RT

By (1)+(2) we get, \(V^{n-1} T=\frac{k}{R}\)

∴ \(\left(\frac{d V}{d T}\right)=-\frac{V}{(n-1) T}=\frac{V}{(1-n) T}\)

According to the first law of thermodynamics,

dQ = \(C_\nu d T+p d V\)

∴ \(\frac{d Q}{d T} =C_\nu+p\left(\frac{d V}{d T}\right)=C_\nu+\frac{p V}{(1-n) T}=C_\nu+\frac{R}{1-n}\)

So, heat capacity, \(C=C_v+\frac{R}{1-n}\)

or, \(1-n=\frac{R}{C-C_v}\)

or, \(n=1-\frac{R}{C-C_v}=\frac{C-\left(C_\nu+R\right)}{C-C_v}=\frac{C-C_p}{C-C_v}\)

(because \(C_p-C_v=R\))

The option 2 is correct.

Question 21. n mol of an ideal gas undergoes a process A→B as shown. The maximum temperature of the gas during the process will be

1. \(\frac{9 p_0 V_0}{4 n R}\)
2. \(\frac{3 p_0 V_0}{2 n R}\)
3. \(\frac{9 p_0 V_0}{2 n R}\)
4. \(\frac{9 p_0 V_0}{n R}\)

Answer:

Equation of straight line AB,

⇒ \(p-p_0=\frac{2 p_0-p}{V_0-2 V_0}\left(V-2 V_0\right)=-\frac{p_0}{V_0}\left(V-2 V_0\right)\)

or, \(p-p_0=-\frac{p_0 V}{V_0}+2 p_0\)

or, p = \(-\frac{p_0 V}{V_0}+3 p_0\)

We know, for n mol of an ideal gas, p V=n R T

∴ \(\frac{n R T}{V}=-\frac{p_0 V}{V_0}+3 p_0\)

or, \(T=-\frac{1}{n R}\left(\frac{p_0 V^2}{V_0}-3 p_0 V\right)\)

At the maximum temperature of the gas \(\frac{d T}{d V}=0\)

∴ \(-\frac{1}{n R}\left(\frac{2 p_0 V}{V_0}-3 p_0\right)=0\)

or, \(\frac{2 p_0 V}{V_0}=3 p_0\)

or, \(V=\frac{3}{2} V_0\)

∴ Maximum temperature,

⇒ \(T_{\text {max }}=-\frac{1}{n R}\left(\frac{9}{4} p_0 V_0-\frac{9}{2} p_0 V_0\right)=\frac{9 p_0 V_0}{4 n R}\)

The option A is correct.

Question 21. C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that C_p-C_v= a for hydrogen gas C_p– C_v= b for nitrogen gas The correct relation between a and b is

1. a = 1/14 b
2. a = b
3. a = 14b
4. a = 28b

Answer:

Given

C_p and C_v are specific heats at constant pressure and constant volume respectively. It is observed that C_p-C_v= a for hydrogen gas C_p– C_v= b for nitrogen ga

Molar specific heat of the gas at constant pressure, \(X_p=M C_p\)

Molar specific heat of the gas at constant volume, \(X_\nu=M C_\nu because X_p-X_\nu=R\)

∴ \(C_p-C_\nu=\frac{R}{M}\)

Then for hydrogen gas, a = R/2 [M_H = 2]

and for nitrogen, b = \(\frac{R}{28}\) (\(M_{\mathrm{N}}=28\))

∴ \(\frac{a}{b}=\frac{28}{2}\) or, a=14 b

The option 3 is correct.

Question 22. A monatomic gas at a pressure p having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V. The final pressure of the gas is (take γ = 5/3)

1. 64p
2. 32p
3. p/64
4. 16p

Answer:

Given

A monatomic gas at a pressure p having a volume V expands isothermally to a volume 2 V and then adiabatically to a volume 16 V.

1st part: p₁V₁ = p₂V₂

∴ pV = p₂ x 2 V or, p₂ = \(\frac{p}{2}\)

2nd part: for adiabatic process \(p V^\gamma=\text { constant } \quad \text { or, } p_1 V_1^\gamma=p_2{ }^{\prime} V_2^\gamma\)

Here, \(p_1=\frac{p}{2}, V_1=2 V, V_2=16 V, \gamma=\frac{5}{3}\)

∴ \(p_2{ }^{\prime}=p_1\left(\frac{V_1}{V_2}\right)^\gamma=\frac{p}{2}\left(\frac{2 V}{16 V}\right)^{5 / 3}=\frac{p}{64}\)

The option 3 is correct.

Question 23. A thermodynamic system undergoes the cyclic process ABCDA as shown. The work done by the system in the cycle is

1. P₀V₀
2. 2p₀V₀
3. \(\frac{p_0 V_0}{2}\)
4. Zero

Answer:

Work done by the system in the cycle = area under the p- V curve and V-axis

= \(\frac{1}{2}\left(2 p_0-p_0\right)\left(2 V_0-V_0\right)+\left[-\left(\frac{1}{2}\right)\left(3 p_0-2 p_0\right)\left(2 V_0-V_0\right)\right]\)

= \(\frac{p_0 V_0}{2}-\frac{p_0 V_0}{2}=0\)

The option 4 is correct.

Question 24. On observing light from three different stars P, Q, and R, it was found that the intensity of the violet color is maximum in the spectrum of P, the intensity of the green color is maximum in the spectrum of R and the intensity of red color in maximum in the spectrum of Q. If T_P, T_Q, and T_R are the respective absolute temperatures of P, Q, and R, then it can be concluded from the above observations that

1. T_P> T_Q> T_R
2. T_P>T_R>T_Q
3. T_P < T_R< T_Q
4. T_P<T_Q<T_R

Answer:

Light of violet color has the shortest wavelength, consequently, it has the highest frequency.

So the temperature of star P will be maximum. Similarly, the light red color has the longest length, and hence star Q has the minimum temperature.

The option 2 is correct.

Question 25. Shows two paths that may be taken by a gas to go from a state A to a state C.

In process, AB, 400 J of heat is added to the system, and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process of AC will be

1. 380 J
2. 500 J
3. 460 J
4. 300 J

Answer:

Change in internal energy over the entire cycle = 0

∴ Q = W

In the case of cycle ABC, Q = W

or, Q_AB + Q_BC+ Q_AC = area of triangle ABC

or, 400 + 100 + Q_CA

= \(\frac{1}{2} \times\left\{(6-2) \times 10^4\right\} \times\left\{(4-2) \times 10^{-3}\right\}\)

= 40

or, \( Q_{C A}=40-400-100=-460 \mathrm{~J}\)

∴ \(Q_{A C}=+460 \mathrm{~J}\)

The option 3 is correct.

Applications of First and Second Laws of Thermodynamics: Short Answers

Question 26. A Carnot engine, having the efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is

1. 100 J
2. 99 J
3. 90 J
4. 1J

Answer:

Given

A Carnot engine, having the efficiency of as heat engine, is used as a refrigerator. If the work done on the system is 10 J

For the heat engine, absorbed energy from the heat reservoir of higher temperature = Q₁ and work obtained, W= 10 J.

Now, \(\eta=\frac{W}{Q_1}=\frac{10}{Q_1}\) given, \(\eta=\frac{1}{10}\) then \(\frac{10}{Q_1}=\frac{1}{10}\) or, \(Q_1=100 \mathrm{~J}\)

So, the energy released at a lower temperature, \(Q_2=Q_1-W=100-10=90 \mathrm{~J}\)

The reverse phenomenon happens in the case of the refrigerator.

So, when the work done on the system is 10 J, then the amount of energy absorbed from the reservoir at a lower temperature = 90 J

The option 3 is correct.

Question 27. One mole of an ideal diatomic gas undergoes n transition from A to H along a path AH as shown,

The change in internal energy of the gas during the transition is

1. 20 KJ
2. -20 KJ
3. 20 J
4. -12KJ

Answer:

For an ideal diatomic gas, \(C_v=\frac{5}{2} R\)

So, change in internal energy,

⇒ \(\Delta U=C_U\left(T_B-T_A\right)=\frac{5}{2} R\left(\frac{p_B V_B}{R}-\frac{p_A V_A}{R}\right)\)

= \(\frac{5}{2}\left(p_B V_B-p_A V_A\right)\)

= \(\frac{5}{2}\left\{\left(2 \times 10^3\right) \times 6-\left(5 \times 10^3\right) \times 4\right\}\)

= \(-\frac{5}{2} \times 8 \times 10^3=-20 \times 10^3 \mathrm{~J}=-20 \mathrm{KJ}\)

The option 2 is correct.

Question 28. A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

1. Compressing the gas through an adiabatic process will require more work to be done
2. Compressing the gas isothermally or dramatically will require the same amount of work
3. Which of the cases (whether compression through isothermal or through an adiabatic process) requires more work will depend upon the atomicity of the gas
4. Compressing the gas isothermally will require more work to be done

Answer:

Given

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half.

Here, the AB curve represents adiabatic compression and the AC curve represents for isothermal compression.

Work done for adiabatic compression = area ABDE

Work done for isothermal compression = area ACDE

As the adiabatic curve is always steeper than the isother¬mal curve, then area ABDE> area ACDE

This means that a higher amount of work will be done in an adiabatic process.

The option 1 is correct.

Question 29. A refrigerator works between 4°C and 30°C. It is required to remove 600 cal of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (take 1 cal = 4,2 J)

1. 23.05 W
2. 236.5 W
3. 2365 W
4. 2.365 W

Answer:

Given

A refrigerator works between 4°C and 30°C. It is required to remove 600 cal of heat every second in order to keep the temperature of the refrigerated space constant.

The coefficient of performance of the refrigerator,

e = \(\frac{T_2}{T_1-T_2}=\frac{277}{303-277}=\frac{277}{26}\)

Here, \(T_2=273+4=277 \mathrm{~K}, T_1=273+30=303 \mathrm{~K}\)

Also, \(e=\frac{Q_2}{W}\) (where, \(Q_2=\) heat absorbed by the refrigerator from colder body

W= work done by refrigerator

or, W = \(\frac{Q_2}{e}=600 \times 4.2 \times \frac{26}{277}=236.53 \mathrm{~J}\)

As work done by the refrigerator is 236.53 ) per second, then the power required is 236.5)3 W.

The option 2 is correct.

Question 30. The volume 1 mol of an ideal gas with the adiabatic exponent y is changed according to the relation V = \(\frac{b}{T}\) where b = constant. The amount of heat absorbed by the gas in the process if the temperature is increased by ΔT will be

1. \(\left(\frac{1-\gamma}{\gamma+1}\right)_{R \Delta T}\)
2. \(\frac{R}{\gamma-1} \Delta T\)
3. \(\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T\)
4. \(\frac{R \Delta T}{\gamma-1}\)

Answer:

Given

The volume 1 mol of an ideal gas with the adiabatic exponent y is changed according to the relation V = \(\frac{b}{T}\) where b = constant.

V = \(\frac{b}{T}\)

∴ dV = \(-\frac{b}{T^2} d T\)

Work done, W = \(\int P d V=-\int \frac{R T}{V} \cdot \frac{b}{T^2} d T\)

= \(-\int \frac{R T}{V} \cdot \frac{V T}{T^2} d T=-R \int d T=-R \Delta T\)

Change in internal energy,

⇒ \(\Delta U=\int C_\nu d T=\frac{R}{\gamma-1} \int d T=\frac{R}{\gamma-1} \Delta T\)

(As \(\frac{R}{\gamma-1}=\frac{C_p-C_v}{\frac{C_p}{C_\nu}-1}=\frac{C_p-C_v}{\frac{C_p-C_\nu}{C_\nu}}=C_v\))

∴ Heat absorbed, Q = \(\Delta U+W=\left(\frac{R}{\gamma-1}-R\right) \Delta T\)

= \(R \times \frac{1-\gamma+1}{\gamma-1} \Delta T=\left(\frac{2-\gamma}{\gamma-1}\right) R \Delta T\)

The option 3 is correct.

Question 31. One mole of a gas obeying the equation of state P(V- b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P-V diagram. Then, the work done is given by

1. \(\frac{1}{2}\left(P_2-P_1\right)\left(V_2+V_1+2 b\right)\)
2. \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)\)
3. \(\frac{1}{2}\left(P_2-P_1\right)\left(V_2-V_1\right)\)
4. \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1+2 b\right)\)

Answer:

Given

One mole of a gas obeying the equation of state P(V- b) = RT is made to expand from a state with coordinates (P₁, V₁) to a state with (P₂, V₂) along a process that is depicted by a straight line on a P-V diagram

Work done = area below straight line 1 → 2

= \(\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)\)

The option 2 is correct.

Question 32. The volume V of a monatomic gas varies with its temperature T, as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

1. 1/3
2. 2/3
3. 2/5
4. 2/7

Answer:

Given

The volume V of a monatomic gas varies with its temperature T, as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it,

Here the curve represents an isobaric process.

In an isobaric process, dQ = nC_pdT….(1)

Degrees of freedom for monatomic gas, f = 3

∴ \(\frac{C_p}{C_v}=\gamma=1+\frac{2}{f}=\frac{5}{3} \text { or, } C_v=\frac{3 C_p}{5}\)

Also, we know, C_p – C_v = R

or, \(C_p-\frac{3 C_p}{5}=R \quad \text { or, } C_p=\frac{5 R}{2}\)

We get from equation (1), \(d Q=n\left(\frac{5}{2} R\right) d T\)

Work is done by gas in this process. dW = pdV = nRdT

∴ \(\frac{d W}{d Q}=\frac{n R d T}{n\left(\frac{5}{2} R\right) d T}=\frac{2}{5}\)

The option 3 is correct.

Question 33. The efficiency of an ideal heat engine working between the freezing point and boiling point of water is

1. 6.25%
2. 20%
3. 26.8%
4. 12.5%

Answer:

Efficiency of an ideal heat engine,

⇒ \(\eta=1-\frac{T_2}{T_1} \quad\left[\begin{array}{c}
T_2=\text { temperature of the sink } \\
T_1=\text { temperature of the source }
\end{array}]\right.\)

or, \(\eta =1-\frac{273}{373}=\frac{100}{373}\)

∴ \(\eta =\frac{100}{373} \times 100 \%\)

= 26.8%

The option 3 is correct

Question 34. A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 10⁵ N · m^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is

1. 42.2J
2. 208.7 J
3. 104.3 J
4. 84.5 J

Answer:

Given

A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 10⁵ N · m^-2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc

Initial volume = 0.1 cm

Final volume = 167.1 cm³

ΔQ = ΔU+ΔW

or, ΔQ = ΔU+pΔV

or, 54×4.2 = AU+1.013 X 10⁵ x (167.1 – 0.1) X 10^-6

or, ΔU = 208.78J

The option 2 is correct

Question 35. Give anyone the difference between a refrigerator and a heat engine. and What type of thermodynamic process is associated with the sudden bursting of an inflated balloon?
Answer:

A refrigerator is a device that transfers heat from lower to higher temperatures, whereas a heat engine is used to convert work into heat.

• In a sudden burst, the air inside the balloon expands to a much higher volume in a very short interval of time. The balloon air does not get sufficient time to exchange heat with its surroundings.
• So the associated thermodynamic process is an adiabatic expansion.

Question 36. Define the isochoric process. What is the work done in such a process?
Answer:

Isochoric process

If a closed thermodynamic system does not change its volume when it exchanges some energy with its environment, the process is known as an isochoric process.

In this process, the volume V = constant; so, dV = 0.

Then, work done, W = ∫pdV = 0, i,e., no work is done in such a process.

Question 36. The molar specific heat capacity at the constant volume of a mono-atomic gas is (3/2)R, where R is the universal gas constant. Find the value of molar specific heat capacity of the gas at constant pressure.
Answer:

Given,

The molar specific heat capacity at the constant volume of a mono-atomic gas is (3/2)R, where R is the universal gas constant.

C_v = \(\frac{3}{2}R\)

From the relation C_p -C_v = R, we have,

∴ \(C_p=C_v+R=\frac{3}{2} R+R=\frac{5}{2} R\)

Question 37. How are these overcome in the second law of thermodynamics? Explain briefly.
Answer:

The 2nd law of thermodynamics overcomes these problems by the introduction of a new variable called entropy (S), which is a property of all thermodynamic systems.

• There are techniques to measure the change in entropy (ΔS) of a system and its surroundings during a process if the entropy increases (ΔS > 0), the 2nd law asserts that the process occurs in nature.
• In the reverse process, the entropy necessarily decreases (ΔS < 0), and that process is not allowed. Moreover, the amount of unavailable energy in a process is directly proportional to ΔS and can be calculated using a simple formula.

Question 38. A refrigerator is maintained (stabilizer kept inside) at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:

Given

A refrigerator is maintained (stabilizer kept inside) at 9°C. If room temperature is 36°C,

Temperature inside the refrigerator, T₁ = 9°C = (273 + 9)K = 282 K

Room temperature, T₂ = 36°C = (273 + 36)K = 309 K

The coefficient of performance of a Carnot refrigerator is

p = \(\frac{T_1}{T_2-T_1}=\frac{282}{309-282}=\frac{282}{27}=10.44\)

This value corresponds to an ideal Carnot refrigerator. All refrigerators in practical use have a lower value of p.

Question 39. State the type of thermodynamic process when

1. Temperature is constant,
2. Volume is constant.

Answer:

1. Isothermal process
2. Isochoric process

Question 40. An ideal gas is compressed at a constant temperature; will its internal energy increase or decrease?
Answer:

As we know, internal energy depends on its temperature. Here temperature is constant, so (dT) = 0, this means that there is no change in internal energy.

First And Second Law Of Thermodynamics Multiple Choice Questions And Answers

Thermodynamics MCQs for Class 11 WBCHSE

Question 1. When a bullet of 6 g mass hits a target at a speed of 400 m · s^-1, 70% of its energy is converted into heat. The value of heat generated is

1. 336 cal
2. 80 cal
3. 3.36 x 10⁵ cal
4. 80000 cal

Answer: 2. 336 cal

Question 2. Water drops from a height of 40 m in a waterfall. If 75% of its energy is converted to heat and absorbed by the water, the rise in temperature of the water will be

1. 0.035°C
2. 0.07°C
3. 0.35°C
4. 0.7°C

Answer: 2. 0.07°C

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. The amount of work done to convert 1 g of ice at 0°C into steam at 100 °C is

1. 756 J
2. 2688 J
3. 3024 J
4. 171.4 J

Answer: 3. 3024 J

Question 4. 169 J energy is required to transform 1g(1cm³) of water into steam at 1 atm pressure. If the latent heat of vaporization of water is 540 cal · g^-1, the volume of that steam will be

1. 1560 cm³
2. 1671 cm³
3. 1571cm³
4. 1600 cm³

Answer: 2. 1671 cm³

Question 5. A man of mass 90 kg gains 10⁵ cal of heat from his food intake. If his digestive ability is 28%, how much height can he climb up to?

1. 1333 m
2. 133.3 m
3. 13.33 m
4. 1.333 m

Answer: 2. 133.3 m

Question 6. Which of the following quantities does not indicate any thermodynamic state of a substance?

1. Volume
2. Temperature
3. Pressure
4. Work

Answer: 4. Work

Question 7. The internal energy of a substance means

1. The kinetic energy of the substance
2. The kinetic energy of the molecules of the substance
3. The sum of its kinetic and potential energies
4. The sum of kinetic and potential energies of the molecules of the substance

Answer: 4. The sum of kinetic and potential energies of the molecules of the substance

First and Second Laws of Thermodynamics MCQs

Question 8. If the average kinetic energy of the molecules of a certain mass of a gas decreases, then

1. The gas becomes hot
2. The gas becomes cold
3. The gas expands
4. The gas contracts

Answer: 2. The gas becomes cold

Question 9. If the volume of a gas of a certain mass changes from 1 L to 0.5 L at a constant pressure of 10⁵ N · m^-2, work done by the gas will be

1. 50000 J
2. -50000 J
3. 50 J
4. -50 J

Answer: 4. -50 J

Question 10. The internal energy of a system is U₁. In a process, work done by the system is W, and heat accepted by the system is Q. At the end of the process, the internal energy of the system is

1. U₁ + Q-W
2. U₁-Q+W
3. U₁ + Q+W
4. U₁ – Q-W

Answer: 1. U₁ + Q-W

Practice MCQs on First Law of Thermodynamics

Question 11. The work done by an ideal gas at constant temperature is 10 J. The amount of heat gained in this process is

1. 10Cal
2. 2.38 cal
3. Zero
4. Data insufficient

Answer: 2. 2.38 cal

Question 12. The work done by an ideal gas at constant pressure is 10 J. The amount of heat gained in this process is

1. 10cal
2. 2.38 cal
3. zero
4. Data insufficient

Answer: 4. Data insufficient

Question 13. If the internal energy U and the work W are expressed in unit of J and the heat is expressed in unit of cal, the first law of thermodynamics will be [here J = Joule’s equivalent]

1. dQ = dU+\(\frac{dW}{J}\)
2. dQ = dU+JdW
3. JdQ = dU+dW
4. \(\frac{dQ}{J}\)

Answer: 3. JdQ = dU+dW

Question 14. C_v = 5/2, for 1 mol of any diatomic ideal ga  of the ratio of the two specific heats is \(\left[\frac{C_p}{C_v}=\gamma\right]\) of the gas is

1. \(\frac{4}{3}\)
2. \(\frac{5}{3}\)
3. \(\frac{7}{3}\)
4. \(\frac{7}{5}\)

Answer: 4. \(\frac{7}{5}\)

Question 15. If R = 2 cal · mol^-1 · °C^-1 and hydrogen is assumed to be an ideal gas, specific heat1 of that gas at constant pressure will be

1. 7 cal g °C^-1
2. 5 cal g °C^-1
3. 3.5 cal · g^{-1 ·} °C^-1
4. 1.25 cal · g^-1 · °C^-1

Answer: 3. 3.5 cal · g^-1 · °C^-1

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 16. Work done becomes zero

1. At constant pressure
2. At constant volume
3. In isothermal process
4. In adiabatic process

Answer: 2. At constant volume

Question 17. The change in internal energy of an ideal gas becomes zero

1. At constant pressure
2. At constant volume
3. In isothermal process
4. In adiabatic process

Answer: 3. In isothermal process

Question 18. The process in which changes in pressure, volume and temperature occur simultaneously is

1. Isobaric
2. Isochoric
3. Isothermal
4. Adiabatic

Answer: 4. Adiabatic

Question 19. It is observed by comparing the specific heats of all solid, liquid, and gaseous substances that

1. Specific heat of water is the highest
2. Specific heats of hydrogen and helium are higher than that of water
3. All gases have specific heat higher than that of water
4. All liquid and gases have specific heats higher than that of water

Answer: 2. Specific heats of hydrogen and helium are higher than that of water

Question 20. In an adiabatic expansion, the change in internal energy of 10 mol of a gas is 100 J. What is the amount of work done by the gas?

1. -100 J
2. 100 J
3. 1000 J
4. -1000 J

Answer: 2. 100 J

Practice MCQs on Second Law of Thermodynamics

Question 21. Which of the following relations does an ideal gas follow in an adiabatic process?

1. pV = RT
2. pV^γ = constant
3. \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\)
4. \(p V^{\gamma-1}\)= constant

Answer: 2. pV^γ = constant

Question 22. The slope of an isothermal curve is always

1. Same as that of an adiabatic curve
2. Greater than that of an adiabatic curve
3. Less than that of an adiabatic curve
4. Not derivable

Answer: 3. Less than that of an adiabatic curve

Question 23. ‘Heat cannot transmit from a body at lower temperature to a body at higher temperature on its own’— which law is this statement derived from?

1. First law of thermodynamics
2. Second law of thermodynamics
3. Law of conservation of momentum
4. Law of conservation of mass

Answer: 2. Second law of thermodynamics

Question 24. A system can go from state A to state B in two different processes 1 and 2. If the change in internal energy in the two cases are ΔU₁ and ΔU₂, respectively, then

1. \(\Delta U_1<\Delta U_2\)
2. \(\Delta U_1>\Delta U_2\)
3. \(\Delta U_1=\Delta U_2\)
4. The relation between \(\Delta U_1 and \Delta U_1\) is uncertain

Answer: 3. \(\Delta U_1=\Delta U_2\)

Question 25. In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas

1. The temperature will decrease
2. The volume will increase
3. The pressure will remain constant
4. The temperature will increase

Answer: 1. The temperature will decrease

Question 26. 5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be T₁, the work done in the process is

1. \(\frac{9}{8} R T_1\)
2. \(\frac{3}{2} R T_1\)
3. \(\frac{15}{8} R T_1\)
4. \(\frac{9}{2} R T_1\)

Answer: 1. \(\frac{9}{8} R T_1\)

Question 27. During the process, shown work done by the system

1. Continuously increases
2. Continuously decreases
3. First increases then decreases
4. First decreases then increases

Answer: 1. Continuously increases

Question 28. When a system is taken from the state i to the state f along the path if, it is found that Q = 50 cal and w= 20 cal. Along the path ibf, Q = 36 cal. W along the path is

1. 6 cal
2. 16 cal
3. 66 cal
4. 14 cal

Answer: 1. 6 cal

Question 29. A Carnot engine, having an efficiency of \(\eta=\frac{1}{10}\) as heat engine is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

1. 99 J
2. 90 J
3. 1J
4. 100 J

Answer: 2. 90 J

Question 30. The efficiency of a Carnot heat engine working between temperatures 127°C and 27°C is

1. \(\frac{27}{127}\)
2. \(\frac{100}{127}\)
3. \(\frac{300}{400}\)
4. \(\frac{100}{400}\)

Answer: 4. \(\frac{100}{400}\)

Sample MCQs on Heat Transfer and Work Done

Question 31. The efficiency of an ideal heat engine is

1. 0%
2. 50%
3. 100%
4. None

Answer: 3. 100%

Question 32. Coefficient of performance of a machine is

1. \(\frac{\text { output }}{\text { input }}\)
2. \(\frac{\text { input }}{\text { output }}\)
3. \(\frac{0}{\text { input }}\)
4. None

Answer: 1. \(\frac{\text { output }}{\text { input }}\)

Question 33. Even Carnot engine cannot give 100% efficiency, because we cannot

1. Prevent radiation
2. Find ideal sources
3. Reach absolute zero temperature
4. Eliminate friction

Answer: 3. Reach absolute zero temperature

In this type of question, more than one option are correct.

Question 34. C_v and C_p denote the molar-specific heat capacities of a gas at constant volume and constant pressure, respectively then,

1. (C_p– C_v) larger for A diatomic ideal gas than for a monatomic ideal gas
2. (C_p + C_v) is larger for a diatomic ideal gas than for a monatomic ideal gas
3. C_p/C_v is a target for a diatomic ideal gas than for a monatomic ideal gas
4. C_p x C_v is larger for a diatomic ideal gas than for a monatomic, ideal gas

Answer:

2. (C_p + C_v) is larger for a diatomic ideal gas than for a monatomic ideal gas

4. C_p x C_v is larger for a diatomic ideal gas than for a monatomic, ideal gas

Key Concepts in Thermodynamics: Multiple Choice Questions

Question 35. Shows the p- V plot of an ideal gas taken through a cycle ABCDA. Part ABC is a semicircle and CDA is half of an ellipse. Then

1. The process during the path A → B is isothermal
2. Heat flows out of the gas during the path B → C → D
3. Work done during the path A → B → C in zero
4. Positive work is done by the gas in the cycle ABCDA

Answer:

1. The process during the path A → B is isothermal

2. Heat flows out of the gas during the path B → C→ D

Question 36. 1 mol of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown. Its pressure at A is p₀. Choose the correct options from the following

1. Internal energies at a and b are the same
2. Work done by the gas in process ab is p₀v₀ ln 4
3. Pressure at C is \(\frac{p_0}{4}\)
4. Temperature at C is \(\frac{t_0}{4}\)

Answer:

1. Internal energies at a and b are the same

3. Pressure at c is \(\frac{p_0}{4}\)

Question 37. In the cyclic process shown, ΔU₁ and ΔU₂ represent the change in internal energy in processes A and B, respectively. If ΔQ is the net heat given to the system in the process and ΔW is the work done by the system in the process. then

1. \(\Delta U_1+\Delta U_2=0\)
2. \(\Delta U_1-\Delta U_2=0\)
3. \(\Delta Q-\Delta W=0\)
4. \(\Delta Q+\Delta W=0\)

Answer:

1. \(\Delta U_1+\Delta U_2=0\)

3. \(\Delta Q-\Delta W=0\)

Question 38. Shows the pV diagrams for a Carnot cycle. In this diagram

1. Curve AB represents the isothermal process and BC adiabatic process
2. Curve AB represents the adiabatic process and BC isothermal process
3. Curve CD represents the isothermal process and DA adiabatic process
4. Curve CD represents the adiabatic process and DA isothermal process

Answer:

2. Curve AB represents the adiabatic process and BC isothermal process

4. Curve CD represents the adiabatic process and DA isothermal process

WBCHSE Class 12 Physics MCQs

Atomic Nucleus Multiple Choice Questions

Question 1. Suppose we consider a large number of containers each containing initially 10000 atoms of radioactive material with a half-life of 1 year. After 1 year,

1. All containers will have 5000 atoms
2. All the containers will contain the same number of
3. Atoms but that number will only be approximately 5000
4. The containers will in general have different numbers of atoms but their average will be close to 5000
5. None ofthe containers can have more than 5000 atoms

Answer: 3. The containers will in general have different numbers of atoms but their average will be close to 5000

Question 2. When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of the atom

1. Do not change for any type of radioactivity
2. Change for α  and β -radioactivity but not for y radioactivity
3. Change for α -radioactivity but not for others
4. Change for β -radioactivity but not for other

Answer: 2. Change for α  and β -radioactivity but not for y radioactivity

Question 3. M_x and M_y denote the atomic masses of the parent and; the daughter nuclei respectively in a radioactive decay. The Q -value for β -decay is Q₁ and that for a β⁺ -decay is Q₂. If m_e denote the mass of an electron, then which of the following statements is correct?

1. Q₁  = (M_x -M_y)c² and Q₂ = (M_x-M_y-2m_e)c²
2. Q₁ = (M_x – M_y)c² and Q₂ = (M_x-M_y)c²
3. Q₁ = (M_x-M_x-2m_e)c² and Q₂ = (M_x-M_y + 2m_e)c²
4. Q₁  = (M_x-M_x + 2m_e)c² and Q₂ = (M_x-M_y+ 2m_e)c²

Answer: 1. Q₁  = (M_x -M_y)c² and Q₂ = (M_x-M_y-2m_e)c²

Read and Learn More Class 12 Physics Multiple Choice Questions

Question 4. Heavy and stable nuclei have more neutrons than protons. This is because the factor

1. Neutrons are heavier than protons
2. Electrostatic force between protons is repulsive
3. Neutrons decay into protons through fi -decay
4. Nuclear forces between neutrons are weaker than those between protons

Answer: 2. Electrostatic force between protons are repulsive

Question 5. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used light nuclei. Heavy nuclei will not serve the purpose because,

1. They will break up
2. Elastic collision of neutrons with heavy nuclei will not slow them down
3. The net weight of the reactor would be unbearably high
4. Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature

Answer:  2. Elastic collision of neutrons with heavy nuclei will not slow them down

WBBSE Class 12 Atomic Nucleus MCQs

Question 6. Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact

1. Nuclear forces have short-range
2. Nuclei are positively charged
3. The original nuclei must be completely ionized before fusion can take place
4. The original nuclei must break up before combining

Answer: 1 and 2

Question 7. The density of the uranium nucleus is approximately

1. 10²⁰ kg. m^-3
2. 10¹⁷ kg. m^–3
3. 10¹³ kg. m^–3
4. 10¹¹ kg. m^{– 3}

Which is the correct option? Given, m_p – 1.67 × 10^-27 kg

Answer: 2. 1 0¹⁷ kg. m^–3

The density of all nuclei is almost the same and its magnitude, ρ ≈ 10¹⁴ g cm^{– 3} = 10¹⁷ kg m^{– 3}

WBCHSE class 12 physics MCQs

Question 8. The approximate value of the density of the uranium nucleus (m_p = 1.67 × 10²⁷ kg) is

1.  10²⁰ kg. m^-3
2.  10¹⁷ kg. m^-3
3.  10¹⁴ kg. m^-3
4.  10¹¹kg.m^-3

Answer: 2.  10¹⁷ kg. m^-3

Question 9. Which of the following is correct?

1. The rest mass of a stable nucleus is less than the sum of the rest masses of the isolated nucleons
2. The rest mass of a stable nucleus is more than the sum of the rest masses of the isolated nucleons
3. In nuclear fusion, energy is emitted due to the combination of two nuclei of comparable masses ( lOOu approx)
4. In nuclear fission, no energy is released due to fragmentation of a very heavy nucleus

Answer:  1. The rest mass of a stable nucleus is less than the sum of the rest masses of the isolated nucleons

Question 10. During the emission of a negative β -particle

1. An electron from the atom is emitted
2. An electron already present inside the nucleus is emitted
3. An electron is emitted due to the disintegration of a neutron inside the nucleus
4. A part of nuclear binding energy is converted into an electron.

Answer:  3. An electron is emitted due to the disintegration of a neutron inside the nucleus

Question 11. Which of the following statements is correct?

1. β-rays and cathode rays are identical.
2. γ -rays are a stream of highly energetic neutrons
3. α -particles are singly charged helium atoms
4. The mass of a proton and that of a neutron are exactly equal

Answer: 1. β-rays and cathode rays are identical.

Question 12. A radioactive nucleus of mass number A, initially at rest, emits an a -particle with a speed v. The recoil speed of the daughter nucleus will be

1. \(\frac{2 v}{A-4}\)
2. \(\frac{2 v}{A+4}\)
3. \(\frac{4 v}{A-4}\)
4. \(\frac{4 v}{A+4}\)

Answer: 3. \(\frac{4 v}{A-4}\)

Question 13. An excited Ne²² nucleus is disintegrated into an unknown nucleus and two α -particles. This unknown nucleus is

1. Nitrogen
2. Carbon
3. Boron
4. Oxygen

Answer: 2. Carbon

Question 14. The half-life of radioiodine I¹³¹ is 8 d. If a sample of I¹³¹ is taken at time t = 0 then it can be said that

1. No nuclear disintegration will occur before t = 4 d
2. No nuclear disintegration will occur before t  8 d
3. All nuclei will be disintegrated in t = 16 d
4. A definite nucleus may be disintegrated at any time after t = 0

Answer: 4. A definite nucleus may be disintegrated at any time after t = 0

WBCHSE class 12 physics MCQs

Question 15. In a freshly prepared radioactive sample, the rate of radia¬ tion is 64 times greater than the safe limit. If its half-life is 2 h then using these safety experiments can be performed safely after

1. 6h
2. 12 h
3. 24 h
4. 128 h

Answer: 2. 12 h

Short Answer Questions on Atomic Nucleus

Question 16. The mean life of a radioactive element is 13 days. Initially, a sample contains 1 g of this element. The mass of the ele¬ ment will be 0.5 g after a time of

1. 13 days
2. 9 days
3. 18.75 days
4. 6.5 days

Answer: 2.  9  days

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 17. The half-life of At²¹⁵ is 100 μs. The time taken for the radioactivity of the sample ofthe element to decay 1/16  th of its initial value is

1. 400 μs
2. 6.3μs
3. 40 μs
4. 300 μs

Answer:   1. 400 μs

Question 18. The half-life of a radioactive substance is 20 min. The approximate time interval (t₂– t₁) between the time t₂ when 2/3 of it has decayed and time t₁ when1/3 of it had decayed is

1. 14 min
2. 20 min
3. 28 min
4. 7 min

Answer: 2. 20 min

Hint: Given, half-life T = \(\frac{\ln 2}{\lambda}\)

T = \(\frac{\ln 2}{\lambda}\) = 20 min

We know that, N = \(N_0 e^{-\lambda t}\)

∴ \(\left(1-\frac{2}{3}\right) N_0=N_0 e^{-\lambda t_2}\)

Or,  \(\frac{1}{3} N_0=N_0 e^{-\lambda t_2}\)……….(1)

Again \(\left(1-\frac{1}{3}\right) N_0=N_0 e^{-\lambda t_1}\)

Or, \(\frac{2}{3} N_0=N_0 e^{-\lambda t_1}\) …………….(2)

Dividing equation (l) by equation (2), we get

⇒ \(\frac{1}{2}=e^{-\lambda\left(t_2-t_1\right)}\)

∴ \(t_2-t_1=\frac{\ln 2}{\lambda}\)

= 20 min

Dual nature of matter and radiation class 12 MCQs 

Question 19. The half-life of radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1: 15 in a sample of a given rock. The age of the rock was estimated to be

1. 200 years
2. 150 years
3. 250 years
4. 100 years

Answer: 2. 150 years

Question 20. A nucleus emits one particle and two γ- particles. The resulting, nucleus is

1. _n-4 Z ^m-6
2. _n Z^m-6
3. _{n  X} ^m-4
4. _{n-2 Y} ^m-4

Answer: 3. _{n X}^m-4

Question 21. Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R . At time t – 0, the number of P species is 4N₀ and that of Q is N₀. The half-life of P (for conversion to R ) is1 min whereas that of Q is 2 min. Initially version of R ) is 1 min whereas that of Q is 2 min. Initially version of R ) is 1 min whereas that of Q is 2 min. Initially R present in the sample would be

1. 2N₀
2. 3N₀
3. \(\frac{9 N_0}{2}\)
4. \(\frac{5 N_0}{2}\)

Answer: 3. \(\frac{9 N_0}{2}\)

Hint: At’ t = 0, the number of nuclei of P and Q respectively 4N₀ and N₀

Let at t = t, the number of nuclei of P and Q are respectively N_P and N_Q

⇒ \(4 N_0\left(\frac{1}{2}\right)^{t / 1}\)

⇒ \(N_Q=N_0\left(\frac{1}{2}\right)^{t / 2}\)

∴N_P = N_Q

∴ \(4 N_0\left(\frac{1}{2}\right)^t=N_0\left(\frac{1}{2}\right)^{t / 2}\)

So, \(\frac{4}{(2)^t}=\frac{1}{2^{t / 2}}\)

t = 4 min

∴ After 4 minutes number of atoms of both types is the same; the Number of atoms after 4 minutes,

⇒ \(\left(4 N_0-\frac{N_0}{4}\right)+\left(N_0-\frac{N_0}{4}\right)=\frac{9 N_0}{2}\)

Common MCQs on Nuclear Structure

Question 22.  Of the following equations which one is the probable nuclear fusion reaction?

1.  ₅C¹³+ ₁H¹ →  ₆C¹⁴+ 4.3 MeV
2. ₆C¹² + ₁H¹ →  ₇N¹³ + 2 MeV
3. ₇N¹⁴+ ₁H¹ →  ₈O¹⁵
4. ₉₂U²³⁵ + ₀n¹ → ₃₄ Xe¹⁴⁰ + ₃₈Sr⁹⁴ + ₀n¹ + ϒ + 200.Mey

Answer: 2.₆C¹² + ₆C¹² + ₁H¹ →  ₇N¹³ + 2 MeV→  ₇N¹³ + 2 MeV

Question 23. In the nuclear reaction \({ }_7^{14} \mathrm{~N}+X \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1^1 \mathrm{H}\) X will be

1.  ₁H¹
2.  ₁H¹
3. ₁H²
4. ₀n¹

Answer: 4. ₀n¹

Question 24. Fast-moving neutrons are retarded

1. By using lead obstacle
2. By passing through water
3. After colliding elastically with heavy nuclei
4. Strong electric fields

Answer: 2. By passing through water

Dual nature of matter and radiation class 12 MCQs 

Question 25. In nuclear fusion

1. A heavy nucleus breaks into two intermediate nuclei and a few high particles
2. A light nucleus breaks due to collision with a thermal neutron
3. A heavy nucleus breaks due to collision with a thermal neutron
4. Two or more light nuclei combine into a heavier nucleus and a few light particles

Answer: 4. Two or more light nuclei combine into a heavier nucleus and a few light particles

Question 26. 4₁H¹ → ₂He⁴ + 2e⁺ + 26 MeV: this is an equation of

1. β-decay
2. γ-decay
3. Fusion
4. Fission

Answer:  3. Fusion

Question 27. The power obtained in a reactor using U²³⁵ disintegration is 1000 kW. The mass decay of U – 235 per hour is

1. 10μg
2. 20μg
3. 40μg
4. 1μg

Answer: 3. 40μg

Question 28. A radioactive isotope XA becomes YA~4 after decay. Which ofthe following radioactive emissions are not possible in this case? ‘

1. α
2. β
3. Meson
4. Positron

Answer: 2,3 and 4

Question 29. A radioactive isotope X^A becomes Y^A-4 after disintegration. Which ofthe following radioactive emissions are not possible in this case?

1. α
2. β
3. Meson
4. Positron

Answer: 1 and 3

Question 30. In the case of a radioactive element which of the following relations are correct where λ = decay constant, T = half-life, and τ = mean life?

1. \(\tau=\frac{1}{\lambda}\)
2. \(\tau=\frac{0.693}{\lambda}\)
3. \(0.6 T\)
4. \(\tau=\frac{T}{0.693}\)

Answer: 1 and 4

Question 31. When α -rays and β -rays are compared as radioactive radiation, it is found that ‘

1. The deflection of β -particles in an electric or magnetic field is comparatively larger
2. The penetration power of β -particles is more
3. The ionization power of β -particles Is more
4. The velocity of β -particles is more

Answer: 1, 2 and 4

Dual nature of matter and radiation class 12 MCQs 

Question 32. The ratio of the mass number of two nuclei is 1: 8, then

1. Ratio of diameter =1:4
2. Ratio of diameter =1:2
3. Ratio of volume =1:8
4. Ratio of volume =1:4

Answer: 2, 3

Question 33. In any nuclear reaction

1. The total number of protons and neutrons remains the same before and after the reaction
2. An increase or decrease in the number of protons is equal to a decrease or increase in the number of neutrons
3. Kinetic energy of the incident particle is approximately 8 MeV or its equivalent
4. Some energy is released if total mass is reduced

Answer: 1, 3 and 4

Question 34. The initial number of radioactive atoms in a radioactive sample is N₀. If after time t the number becomes N, then N = N₀ e^-λt, where λ is known as the decay constant ofthe element. The time in which the number of radioactive atoms becomes half of its initial number is called the half-life (T) of the element. The time in which the number of atoms falls to 1/e times its initial number is the mean life (τ) ofthe element. The product λN is the activity (A) ofthe radioactive sample when the number of atoms is N. The SI unit of activity is bequerel (Bq); where lBq = 1decay. s^-1, and Avogadro’s number, N = 6.023 × 10²³.

1. The half-life of iodine-131 is 8d. Its decay constant (in SI)

1. 10^-6
2. 1.45 × 10^-6
3. 2 × 10^-6
4. 2.9  × 10^-6

Answer: 1. 10^-6

2. The half-life of iodine-131 is 8d. Its mean life (in SI ) is

1. 4.79 × 10⁵
2. 6.912× 10⁵
3. 9.974× 10⁵
4. 22. 96 × 10⁵

Answer: 3. 9.974× 10⁵

3. The half-life of Iodine 131 is 8d. What is the activity Bq) of 1 g of iodine?

1. 2.3 × 10¹⁵
2. 4.6 × 10¹⁵
3. 6.9 × 10¹⁵
4. 9.2 × 10¹⁵

Answer: 2. 4.6 × 10¹⁵

Class 12 physics dual nature questions 

4.  in the equation above After how many days the activity of iodine-131 will be \(\frac{1}{16}\)th of its initial value

1. 24 data
2. 32 data
3. 40 data
4. 48 data

Answer: 2. 32 data

5. In the question above, what is the ratio of the activity of sodium- 24 to that of iodine-131 (half-life of sodium- 24 is 15h)?

1. \(\frac{1}{70}\)
2. \(\frac{1}{7}\)
3. 7
4. 70

Answer: 4. 70

Question 35. For the radioactive nuclei that undergo either a or /S decay, which one of the following cannot occur?

1. Isobar of the original nucleus is produced
2. Isotope of the original nucleus is produced
3. Nuclei with higher atomic numbers than that of the original nucleus is produced
4. Nuclei with lower atomic number than that of the

Answer: 2. Isotope of the original nucleus is produced

Practice MCQs on Nuclear Forces

Question 36. Radon-222 has a half-life of 3.8 days. If one starts with 0.064 kg of radon-222 the quantity of radon-222 left after 19 days
will be

1. 0.002 kg
2. 0.032 kg
3. 0.062 kg
4. 0.024 kg

Answer: 1. 0.002 kg

⇒ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^{t / T}\)

Or, \(N=N_0\left(\frac{1}{2}\right)^{19 / 3.8}\)

= \(0.064 \times \frac{1}{32}\)

= 0.002 kg

Question 37. If the half-life of a radioactive nucleus is 3 days, nearly what fraction of the initial number of nuclei will decay on the 3rd day? (given, \(\sqrt[3]{0.25}\) = 0.63 )

1. 0.63
2. 0.37
3. 0.5
4. 0.13

Answer: 4. 0.13

We know, in case of radioactive decay

N = \(N_0 e^{-\lambda t}\)

Again = \(\lambda=\frac{\ln 2}{T_{1 / 2}}\)

Given T = 3 day

∴ \(\lambda=\frac{\ln 2}{3}\)

The fraction ofthe initial number of nuclei will decay on the 3rd day

= \(\frac{N_0 e^{-\frac{\ln 2}{3} \times 2}-N_0 e^{-\frac{\ln 2}{3} \times 3}}{N_0}=e^{-\frac{2 \ln 2}{3}}-e^{-\frac{3 \ln 2}{3}}\)

= \(2^{-\frac{2}{3}}-2^{-1}\)

= 0.13

Class 12 physics dual nature questions 

Question 38. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have an equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be

1. 1:16
2. 4:1
3. 1:4
4. 5:4

Answer: 4. 5:4

80 minutes = 4 half-lives of A = 2 half-lives of B

Let the initial number of nuclei in each sample be N.

Number of undecayed nuclides of element A after 80 minutes \(\)

Number of A nuclides decayed = \(\frac{N}{2^2}\)

Number of undecayed nuclides of element B after 80 minutes = \(\frac{N}{2^2}\)

Number of B nuclides decayed \(\frac{3}{4}\)

Required ratio = \(\frac{15 / 16}{3 / 4}=\frac{5}{4}\)

Question 39. A radioactive nucleus A with a half-life of T, decays into a nucleus B. At t = 0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by

1. t = \(\frac{T}{2} \frac{\log 2}{\log 1.3}\)
2. t = \(T \frac{\log 1.3}{\log 2}\)
3. t = \(T \log (1.3)\)
4. t = \(\frac{T}{\log (1.3)}\)

Answer:  2. t = \(T \frac{\log 1.3}{\log 2}\)

After time t, the number of nuclei of A

⇒ \(N_B=N_0-N_A=N_0\left(1-e^{-\lambda t}\right)\)

⇒  \(\frac{N_B}{N_A}\) = 0.3

⇒  \(\frac{1-e^{-\lambda t}}{e^{-\lambda t}}\) = 0.3

t = In 1.3

\(\left(\frac{\ln 2}{T}\right) t\)= In 1.3

t = \(T \frac{\ln 1.3}{\ln 2}=T \frac{\log 1.3}{\log 2}\)

Question 40. The binding energy per nucleon of ₃Li⁷ and ₂He⁴  nuclei are 5.60MeV and 7.06MeV respectively. In the nuclear reaction \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\) the value of energy Q released is

1. 19.6 MeV
2. – 2.4 meV
3. 8.4 MeV
4. 17.3 MeV

Answer: 4. 17.3 MeV

Binding energy of ₂He⁴ = 4 ×7.06 MeV

Binding energy of ₃Li⁷ = 7 × 5.60 MeV

The nuclear equation is

⇒ \({ }_3^7 \mathrm{Li}+{ }_1^1 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+\mathrm{Q}\)

7 × 5.60 = 4 × 7.06 + 4 × 7.06 + Q

or, Q = 56.48- 39.20 = 17.28 MeV

Question 41. A radioisotope X with a half-life of 1.4 × 10⁹ years decays to stable Y. A sample of the rock from a cave was found to contain X and Y in a ratio of 1:7. The age of the rock is

1. 1.96 × 10⁹ years
2. 3.92 × 10⁹ y
3. 3.92 × 10⁹ y
4. 4.20 × 10⁹ years

Answer:  3. 3.92 × 10⁹ y

Suppose N atoms out of N₀ atoms of element X disintegrate to element Y in time t

∴ \(\frac{N}{N_0-N}=\frac{1}{7} \quad \text { or, } \frac{N}{N_0}=\frac{1}{8}\)

∴ \(\frac{N}{N_0}=\left(\frac{1}{2}\right)^3\)

∴  Time taken for this disintegration is 3 times the half-life.

t = 3 × 1.4 × 10⁹ = 4.2 × 10⁹ years

Conceptual Questions on Radioactivity

Question 42. If the radius of the ₁₃Li²⁷ nucleus is taken to be RM, then the 125 radius of the ₅₃Te¹²⁵ nucleus is nearly

1. \(\left(\frac{53}{13}\right)^{1 / 3} R_{\mathrm{Al}}\)
2. \(\frac{5}{3} R_{\mathrm{Al}}\)
3. \(\frac{3}{5} R_{\mathrm{Al}}\)
4. \(\left(\frac{13}{53}\right)^{1 / 3} R_{\mathrm{Al}}\)

Answer: 2. \(\frac{5}{3} R_{\mathrm{Al}}\)

We know R = \(r_0 \mathrm{~A}^{1 / 3}\)

Then R Al = \(r_0(27)^{1 / 3}\) = 3rd

= \(R_{\mathrm{Te}}=r_0(125)^{1 / 3}\) = 5r

= \(\frac{5}{3} \cdot 3 r_0=\frac{5}{3} R_{\mathrm{Al}}\)

Question 43. The energy liberated per nuclear fission 10zo fissions occur per second the amount of power produced will be

1. 2 × 10²² W
2. 32 × 10⁸ W
3. 16 × 10⁸ W
4. 5 × 10¹¹ W

Answer: 2. 32 × 10⁸ W

Power = Energy liberated per nuclear fission x number of fissions per second

= 200 × 10²⁰ MeV/s

= (200 × 10⁶ × 1. 6 × 10^-19) × 10²⁰ J/s

= 3.2 × 10⁹

= 32 × 10⁸W

Question 44. For a radioactive material, the half-life is 10 minutes. If initially there are 600 nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

1. 3
2. 10
3. 20
4. 15

Answer: 3. 20

⇒ \(\left(\frac{N}{N_0}\right)=\left(\frac{1}{2}\right)^{t / T}\)

⇒ \(\frac{600-450}{600}=\left(\frac{1}{2}\right)^{t / 10}\)

or, t/10 = 2

t = 20 min

Photoelectric effect multiple choice questions 

Question 45. A radioactive element emits 2 or -particles and 3 0 – particles. The values of atomic number (Z) and mass number (A) of the new element will be
Answer:

1. (A +5),(Z-1)
2. (A – 5) ,(Z+1)
3. (A -8),(Z-1)
4. (A – 8) ,(Z+1)

Answer: 3. (A -8),(Z-1)

The mass number after emission of 2 or -particles

= A – (2 × 4) = (A – 8)

The atomic number after emission of 2 α -particles and 3 β   particles

Z- (2 × 2) + (3 × 1)  = Z-1